Chapter 14 : Equation of a Line - Selina Solutions for Class 10 Maths ICSE

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Chapter 14 - Equation of a Line Excercise Ex. 14(C)

Question 1

Find the equation of a line whose:

y-intercept = 2 and slope = 3.

Solution 1

Given, y-intercept = c = 2 and slope = m = 3.

Substituting the values of c and m in the equation y = mx + c, we get,

y = 3x + 2, which is the required equation.

Question 2

Find the equation of a line whose:

y-intercept = -1 and inclination = 45o.

Solution 2

Given, y-intercept = c = -1 and inclination = 45o.

Slope = m = tan 45o = 1

Substituting the values of c and m in the equation y = mx + c, we get,

y = x - 1, which is the required equation.

Question 3

Find the equation of the line whose slope is and which passes through (-3, 4).

Solution 3

Given, slope =

The equation passes through (-3, 4) = (x1, y1)

Substituting the values in y - y1 = m(x - x1), we get,

y - 4 = (x + 3)

3y - 12 = -4x - 12

4x + 3y = 0, which is the required equation.

Question 4

Find the equation of a line which passes through (5, 4) and makes an angle of 60o with the positive direction of the x-axis.

Solution 4

Slope of the line = tan 60o =

The line passes through the point (5, 4) = (x1, y1)

Substituting the values in y - y1 = m(x - x1), we get,

y - 4 = (x - 5)

y - 4 = x - 5

y =x + 4 - 5, which is the required equation.

Question 5

Find the equation of the line passing through:

(i) (0, 1) and (1, 2) (ii) (-1, -4) and (3, 0)

Solution 5

(i) Let (0, 1) = (x1, y1) and (1, 2) = (x2, y2)

The required equation of the line is given by:

y - y1 = m(x - x1)

y - 1 = 1(x - 0)

y - 1 = x

y = x + 1

(ii) Let (-1, -4) = (x1, y1) and (3, 0) = (x2, y2)

The required equation of the line is given by:

y - y1 = m(x - x1)

y + 4 = 1(x + 1)

y + 4 = x + 1

y = x - 3

Question 6

The co-ordinates of two points P and Q are (2, 6) and (-3, 5) respectively. Find:

(i) the gradient of PQ;

(ii) the equation of PQ;

(iii) the co-ordinates of the point where PQ intersects the x-axis.

Solution 6

Given, co-ordinates of two points P and Q are (2, 6) and (-3, 5) respectively.

(i) Gradient of PQ =

(ii) The equation of the line PQ is given by:

y - y1 = m(x - x1)

y - 6 = (x - 2)

5y - 30 = x - 2

5y = x + 28

(iii) Let the line PQ intersects the x-axis at point A (x, 0).

Putting y = 0 in the equation of the line PQ, we get,

0 = x + 28

x = -28

Thus, the co-ordinates of the point where PQ intersects the x-axis are A (-28, 0).

Question 7

The co-ordinates of two points A and B are (-3, 4) and (2, -1). Find:

(i) the equation of AB;

(ii) the co-ordinates of the point where the line AB intersects the y-axis.

Solution 7

(i) Given, co-ordinates of two points A and B are (-3, 4) and (2, -1).

Slope =

The equation of the line AB is given by:

y - y1 = m(x - x1)

y + 1 = -1(x - 2)

y + 1 = -x + 2

x + y = 1

(ii) Let the line AB intersects the y-axis at point (0, y).

Putting x = 0 in the equation of the line, we get,

0 + y = 1

y = 1

Thus, the co-ordinates of the point where the line AB intersects the y-axis are (0, 1).

Question 8

The figure given below shows two straight lines AB and CD intersecting each other at point P (3, 4). Find the equation of AB and CD.

Solution 8

Slope of line AB = tan 45o = 1

The line AB passes through P (3, 4). So, the equation of the line AB is given by:

y - y1 = m(x - x1)

y - 4 = 1(x - 3)

y - 4 = x - 3

y = x + 1

Slope of line CD = tan 60o =

The line CD passes through P (3, 4). So, the equation of the line CD is given by:

y - y1 = m(x - x1)

y - 4 = (x - 3)

y - 4 = x - 3

y = x + 4 - 3

Question 9

In ΔABC, A = (3, 5), B = (7, 8) and C = (1, -10). Find the equation of the median through A.

Solution 9

Question 10

The following figure shows a parallelogram ABCD whose side AB is parallel to the x-axis, A = 60o and vertex C = (7, 5). Find the equations of BC and CD.

Solution 10

Since, ABCD is a parallelogram,

B = 180o - 60o = 120o

Slope of BC = tan 120o = tan (90o + 30o) = cot30o =

Equation of the line BC is given by:

y - y1 = m(x - x1)

y - 5 = (x - 7)

y - 5 = x - 7

y = x + 5 - 7

Since, CD || AB and AB || x-axis, slope of CD = Slope of AB = 0

Equation of the line CD is given by:

y - y1 = m(x - x1)

y - 5 = 0(x - 7)

y = 5

Question 11

Find the equation of the straight line passing through origin and the point of intersection of the lines x + 2y = 7 and x - y = 4.

Solution 11

The given equations are:


x + 2y = 7 ....(1)


x - y = 4 ....(2)

 

Subtracting (2) from (1), we get,


3y = 3


y = 1

 

From (2), x = 4 + y = 4 + 1 = 5

 

The required line passes through (0, 0) and (5, 1).


S l o p e space o f space t h e space l i n e equals fraction numerator 1 minus 0 over denominator 5 minus 0 end fraction equals 1 fifth

 

Required equation of the line is given by:


y minus y subscript 1 equals m open parentheses x minus x subscript 1 close parentheses
rightwards double arrow y minus 0 equals 1 fifth open parentheses x minus 0 close parentheses
rightwards double arrow 5 y equals x
rightwards double arrow x minus 5 y equals 0

Question 12

In triangle ABC, the co-ordinates of vertices A, B and C are (4, 7), (-2, 3) and (0, 1) respectively. Find the equation of median through vertex A.

Also, find the equation of the line through vertex B and parallel to AC.

Solution 12

Given, the co-ordinates of vertices A, B and C of a triangle ABC are (4, 7), (-2, 3) and (0, 1) respectively.

Let AD be the median through vertex A.

Co-ordinates of the point D are

Slope of AD =

The equation of the median AD is given by:

y - y1 = m(x - x1)

y - 2 = 1(x + 1)

y - 2 = x + 1

y = x + 3

The slope of the line which is parallel to line AC will be equal to the slope of AC.

Slope of AC =

The equation of the line which is parallel to AC and passes through B is given by:

y - 3 = (x + 2)

2y - 6 = 3x + 6

2y = 3x + 12

Question 13

A, B and C have co-ordinates (0, 3), (4, 4) and (8, 0) respectively. Find the equation of the line through A and perpendicular to BC.

Solution 13

Slope of BC =

Slope of line perpendicular to BC =

The equation of the line through A and perpendicular to BC is given by:

y - y1 = m(x - x1)

y - 3 = 1(x - 0)

y - 3 = x

y = x + 3

Question 14

Find the equation of the perpendicular dropped from the point (-1, 2) onto the line joining the points (1, 4) and (2, 3).

Solution 14

Let A = (1, 4), B = (2, 3), and C = (-1, 2).

Slope of AB =

Slope of equation perpendicular to AB =

The equation of the perpendicular drawn through C onto AB is given by:

y - y1 = m(x - x1)

y - 2 = 1(x + 1)

y - 2 = x + 1

y = x + 3

Question 15

Find the equation of the line, whose:

(i) x-intercept = 5 and y-intercept = 3

(ii) x-intercept = -4 and y-intercept = 6

(iii) x-intercept = -8 and y-intercept = -4

Solution 15

(i) When x-intercept = 5, corresponding point on x-axis is (5, 0)

When y-intercept = 3, corresponding point on y-axis is (0, 3).

Let (x1, y1) = (5, 0) and (x2, y2) = (0, 3)

Slope =

 

The required equation is:

y - y1 = m(x - x1)

y - 0 = (x - 5)

5y = -3x + 15

3x + 5y = 15

 

(ii) When x-intercept = -4, corresponding point on x-axis is (-4, 0)

When y-intercept = 6, corresponding point on y-axis is (0, 6).

Let (x1, y1) = (-4, 0) and (x2, y2) = (0, 6)

Slope =

 

The required equation is:

y - y1 = m(x - x1)

y - 0 = (x + 4)

2y = 3x + 12

 

(iii) When x-intercept = -8, corresponding point on x-axis is (-8, 0)

When y-intercept = -4, corresponding point on y-axis is (0, -4).

Let (x1, y1) = (-8, 0) and (x2, y2) = (0, -4)

Slope =

 

The required equation is:

y - y1 = m(x - x1)

y - 0 = (x + 8)

2y = -x - 8

x + 2y + 8 = 0

Question 16

Find the equation of the line whose slope is and x-intercept is 6.

Solution 16

Since, x-intercept is 6, so the corresponding point on x-axis is (6, 0).

Slope = m =

Required equation of the line is given by:

y - y1 = m(x - x1)

y - 0 = (x - 6)

6y = -5x + 30

5x + 6y = 30

Question 17

Find the equation of the line with x-intercept 5 and a point on it (-3, 2).

Solution 17

Since, x-intercept is 5, so the corresponding point on x-axis is (5, 0).

The line also passes through (-3, 2).

Slope of the line =

Required equation of the line is given by:

y - y1 = m(x - x1)

y - 0 = (x - 5)

4y = -x + 5

x + 4y = 5

Question 18

Find the equation of the line through (1, 3) and making an intercept of 5 on the y-axis.

Solution 18

Since, y-intercept = 5, so the corresponding point on y-axis is (0, 5).

The line passes through (1, 3).

Slope of the line =

Required equation of the line is given by:

y - y1 = m(x - x1)

y - 5 = -2(x - 0)

y - 5 = -2x

2x + y = 5

Question 19

Find the equations of the lines passing through point (-2, 0) and equally inclined to the co-ordinate axis.

Solution 19

Let AB and CD be two equally inclined lines.

For line AB:

Slope = m = tan 45o = 1

(x1, y1) = (-2, 0)

Equation of the line AB is:

y - y1 = m(x - x1)

y - 0 = 1(x + 2)

y = x + 2

For line CD:

Slope = m = tan (-45o) = -1

(x1, y1) = (-2, 0)

Equation of the line CD is:

y - y1 = m(x - x1)

y - 0 = -1(x + 2)

y = -x - 2

x + y + 2 = 0

Question 20

The line through P(5, 3) intersects y-axis at Q.

(i) Write the slope of the line.

(ii) Write the equation of the line.

(iii) Find the co-ordinates of Q.


Solution 20

(i)

 

 

(ii)

 

 

(iii)

Question 21

Write down the equation of the line whose gradient is and which passes through point P, where P divides the line segment joining A (4, -8) and B (12, 0) in the ratio 3: 1.

Solution 21

Given, P divides the line segment joining A (4, -8) and B (12, 0) in the ratio 3: 1.

Co-ordinates of point P are

Slope = m = (Given)

Thus, the required equation of the line is

y - y1 = m(x - x1)

y + 2 = (x - 10)

5y + 10 = -2x + 20

2x + 5y = 10

Question 22

A (1, 4), B (3, 2) and C (7, 5) are vertices of a triangle ABC, Find:

(i) the co-ordinates of the centroid of triangle ABC.

(ii) the equation of a line, through the centroid and parallel to AB.

Solution 22

(i) Co-ordinates of the centroid of triangle ABC are

 

(ii) Slope of AB =

Slope of the line parallel to AB = Slope of AB = -1

 

Thus, the required equation of the line is

y - y1 = m(x - x1)

Question 23

A (7, -1), B (4, 1) and C (-3, 4) are the vertices of a triangle ABC. Find the equation of a line through the vertex B and the point P in AC; such that AP: CP = 2: 3.

Solution 23

Given, AP: CP = 2: 3

Co-ordinates of P are

Slope of BP =

Required equation of the line passing through points B and P is

y - y1 = m(x - x1)

y - 1 = 0(x - 3)

y = 1

Chapter 14 - Equation of a Line Excercise Ex. 14(D)

Question 1

Find the slope and y-intercept of the line:

(i) y = 4

(ii) ax - by = 0

(iii) 3x - 4y = 5

Solution 1

(i) y = 4

Comparing this equation with y = mx + c, we have:

Slope = m = 0

y-intercept = c = 4

(ii) ax - by = 0 by = ax y =

Comparing this equation with y = mx + c, we have:

Slope = m =

y-intercept = c = 0

(iii) 3x - 4y = 5

Comparing this equation with y = mx + c, we have:

Slope = m =

y-intercept = c =

Question 2

The equation of a line x - y = 4. Find its slope and y-intercept. Also, find its inclination.

Solution 2

Given equation of a line is x - y = 4

y = x - 4

Comparing this equation with y = mx + c. We have:

Slope = m = 1

y-intercept = c = -4

Let the inclination be .

Slope = 1 = tan = tan 45o

Question 3

(i) Is the line 3x + 4y + 7 = 0 perpendicular to the line 28x - 21y + 50 = 0?

(ii) Is the line x - 3y = 4 perpendicular to the line 3x - y = 7?

(iii) Is the line 3x + 2y = 5 parallel to the line x + 2y = 1?

(iv) Determine x so that the slope of the line through (1, 4) and (x, 2) is 2.

Solution 3

(i) 3x + 4y + 7 = 0

Slope of this line =

28x - 21y + 50 = 0

Slope of this line =

Since, product of slopes of the two lines = -1, the lines are perpendicular to each other.

(ii) x - 3y = 4

3y = x - 4

y =

Slope of this line =

3x - y = 7

y = 3x - 7

Slope of this line = 3

Product of slopes of the two lines = 1 -1

So, the lines are not perpendicular to each other.

(iii) 3x + 2y = 5

2y = -3x + 5

y =

Slope of this line =

x + 2y = 1

2y = -x + 1

y =

Slope of this line =

Product of slopes of the two lines = 3 -1

So, the lines are not perpendicular to each other.

(iv) Given, the slope of the line through (1, 4) and (x, 2) is 2.

Question 4

Find the slope of the line which is parallel to:

(i) x + 2y + 3 = 0 (ii)

Solution 4

(i) x + 2y + 3 = 0

2y = -x - 3

y =

Slope of this line =

Slope of the line which is parallel to the given line = Slope of the given line =

(ii)

Slope of this line =

Slope of the line which is parallel to the given line = Slope of the given line =

Question 5

Find the slope of the line which is perpendicular to:

(i) (ii)

Solution 5

(i)

Slope of this line = 2

Slope of the line which is perpendicular to the given line =

(ii)

Slope of this line =

Slope of the line which is perpendicular to the given line =

Question 6

(i) Lines 2x - by + 3 = 0 and ax + 3y = 2 are parallel to each other. Find the relation connecting a and b.

(ii) Lines mx + 3y + 7 = 0 and 5x - ny - 3 = 0 are perpendicular to each other. Find the relation connecting m and n.

Solution 6

(i) 2x - by + 3 = 0

by = 2x + 3

y =

Slope of this line =

ax + 3y = 2

3y = -ax + 2

y =

Slope of this line =

Since, the lines are parallel, so the slopes of the two lines are equal.


(ii) mx + 3y + 7 = 0

3y = -mx - 7

y =

Slope of this line =

5x - ny - 3 = 0

ny = 5x - 3

y =

Slope of this line =

Since, the lines are perpendicular; the product of their slopes is -1.

Question 7

Find the value of p if the lines, whose equations are 2x - y + 5 = 0 and px + 3y = 4 are perpendicular to each other.

Solution 7

2x - y + 5 = 0

y = 2x + 5

Slope of this line = 2

px + 3y = 4

3y = -px + 4

y =

Slope of this line =

Since, the lines are perpendicular to each other, the product of the slopes is -1.

Question 8

The equation of a line AB is 2x - 2y + 3 = 0.

(i) Find the slope of the line AB.

(ii) Calculate the angle that the line AB makes with the positive direction of the x-axis.

Solution 8

(i) 2x - 2y + 3 = 0

2y = 2x + 3

y = x +

Slope of the line AB = 1

(ii) Required angle =

Slope = tan = 1 = tan 45o

= 45o

Question 9

The lines represented by 4x + 3y = 9 and px - 6y + 3 = 0 are parallel. Find the value of p.

Solution 9

4x + 3y = 9

3y = -4x + 9

y = + 3

Slope of this line =

px - 6y + 3 = 0

6y = px + 3

y =

Slope of this line =

Since, the lines are parallel, their slopes will be equal.

Question 10

If the lines y = 3x + 7 and 2y + px = 3 are perpendicular to each other, find the value of p.

Solution 10

y = 3x + 7

Slope of this line = 3

2y + px = 3

2y = -px + 3

y =

Slope of this line =

Since, the lines are perpendicular to each other, the product of their slopes is -1.

Question 11

The line through A(-2,3) and B(4,b) is perpendicular to the line 2x - 4y =5. Find the value of b.

Solution 11

 

 

 

Question 12

Find the equation of the line through (-5, 7) and parallel to:

(i) x-axis (ii) y-axis

Solution 12

(i) The slope of the line parallel to x-axis is 0.

(x1, y1) = (-5, 7)

Required equation of the line is

y - y1 = m(x - x1)

y - 7 = 0(x + 5)

y = 7

(ii) The slope of the line parallel to y-axis is not defined.

That is slope of the line is and hence the given line is parallel to y-axis.

(x1, y1) = (-5, 7)

Required equation of the line is

x - x1 =0

x + 5=0

Question 13

(i) Find the equation of the line passing through (5, -3) and parallel to x - 3y = 4.

(ii) Find the equation of the line parallel to the line 3x + 2y = 8 and passing through the point (0, 1).

Solution 13

(i) x - 3y = 4

3y = x - 4

Slope of this line =

Slope of a line parallel to this line =

Required equation of the line passing through (5, -3) is

y - y1 = m(x - x1)

y + 3 = (x - 5)

3y + 9 = x - 5

x - 3y - 14 = 0

(ii) 2y = -3x + 8

Or y =

Slope of given line =

Since the required line is parallel to given straight line.

Slope of required line (m) =

Now the equation of the required line is given by:

y - y1 = m(x - x1)

y - 1 =

2y - 2 = -3x

3x + 2y = 2

Question 14

Find the equation of the line passing through (-2, 1) and perpendicular to 4x + 5y = 6.

Solution 14

4x + 5y = 6

5y = -4x + 6

y =

Slope of this line =

The required line is perpendicular to the line 4x + 5y = 6.

The required equation of the line is given by

y - y1 = m(x - x1)

y - 1 = (x + 2)

4y - 4 = 5x + 10

5x - 4y + 14 = 0

Question 15

Find the equation of the perpendicular bisector of the line segment obtained on joining the points (6, -3) and (0, 3).

Solution 15

Let A = (6, -3) and B = (0, 3).

We know the perpendicular bisector of a line is perpendicular to the line and it bisects the line, that it, it passes through the mid-point of the line.

Co-ordinates of the mid-point of AB are

Thus, the required line passes through (3, 0).

Slope of AB =

Slope of the required line =

Thus, the equation of the required line is given by:

y - y1 = m(x - x1)

y - 0 = 1(x - 3)

y = x - 3

Question 16

In the following diagram, write down:

(i) the co-ordinates of the points A, B and C.

(ii) the equation of the line through A and parallel to BC.

Solution 16

(i) The co-ordinates of points A, B and C are (2, 3), (-1, 2) and (3, 0) respectively.

(ii) Slope of BC =

Slope of a line parallel to BC = Slope of BC =

Required equation of a line passing through A and parallel to BC is given by

y - y1 = m(x - x1)

y - 3 = (x - 2)

2y - 6 = -x + 2

X + 2y = 8

Question 17

B (-5, 6) and D (1, 4) are the vertices of rhombus ABCD. Find the equation of diagonal BD and of diagonal AC.

Solution 17

We know that in a rhombus, diagonals bisect each other at right angle.

Let O be the point of intersection of the diagonals AC and BD.

Co-ordinates of O are

Slope of BD =

For line BD:

Slope = m = , (x1, y1) = (-5, 6)

Equation of the line BD is

y - y1 = m(x - x1)

y - 6 = (x + 5)

3y - 18 = -x - 5

x + 3y = 13

For line AC:

Slope = m = , (x1, y1) = (-2, 5)

Equation of the line AC is

y - y1 = m(x - x1)

y - 5 = 3(x + 2)

y - 5 = 3x + 6

y = 3x + 11

Question 18

A = (7, -2) and C = (-1, -6) are the vertices of square ABCD. Find the equations of diagonal BD and of diagonal AC.

Solution 18

We know that in a square, diagonals bisect each other at right angle.

Let O be the point of intersection of the diagonals AC and BD.

Co-ordinates of O are

Slope of AC =

For line AC:

Slope = m = , (x1, y1) = (7, -2)

Equation of the line AC is

y - y1 = m(x - x1)

y + 2 = (x - 7)

2y + 4 = x - 7

2y = x - 11

For line BD:

Slope = m = , (x1, y1) = (3, -4)

Equation of the line BD is

y - y1 = m(x - x1)

y + 4 = -2(x - 3)

y + 4 = -2x + 6

2x + y = 2

Question 19

A (1, -5), B (2, 2) and C (-2, 4) are the vertices of triangle ABC, find the equation of:

(i) the median of the triangle through A.

(ii) the altitude of the triangle through B.

(iii) the line through C and parallel to AB.

Solution 19

(i) We know the median through A will pass through the mid-point of BC. Let AD be the median through A.

Co-ordinates of the mid-point of BC, i.e., D are

Slope of AD =

Equation of the median AD is

y - 3 = -8(x - 0)

8x + y = 3

(ii) Let BE be the altitude of the triangle through B.

Slope of AC =

Slope of BE =

Equation of altitude BE is

y - 2 = (x - 2)

3y - 6 = x - 2

3y = x + 4

(iii) Slope of AB =

Slope of the line parallel to AB = Slope of AB = 7

So, the equation of the line passing through C and parallel to AB is

y - 4 = 7(x + 2)

y - 4 = 7x + 14

y = 7x + 18

Question 20

(i) Write down the equation of the line AB, through (3, 2) and perpendicular to the line 2y = 3x + 5.

(ii) AB meets the x-axis at A and the y-axis at B. Write down the co-ordinates of A and B. Calculate the area of triangle OAB, where O is the origin.

Solution 20

(i) 2y = 3x + 5

Slope of this line =

Slope of the line AB =

(x1, y1) = (3, 2)

The required equation of the line AB is

y - y1 = m(x - x1)

y - 2 = (x - 3)

3y - 6 = -2x + 6

2x + 3y = 12

(ii) For the point A (the point on x-axis), the value of y = 0.

2x + 3y = 12 2x = 12 x = 6

Co-ordinates of point A are (6, 0).

For the point B (the point on y-axis), the value of x = 0.

2x + 3y = 12 3y = 12 y = 4

Co-ordinates of point B are (0, 4).

Area of OAB = OA OB = 6 4 = 12 sq units

Question 21

The line 4x - 3y + 12 = 0 meets the x-axis at A. Write the co-ordinates of A.

Determine the equation of the line through A and perpendicular to 4x - 3y + 12 = 0.

Solution 21

For the point A (the point on x-axis), the value of y = 0.

4x - 3y + 12 = 0 4x = -12 x = -3

Co-ordinates of point A are (-3, 0).

Here, (x1, y1) = (-3, 0)

The given line is 4x - 3y + 12 = 0

3y = 4x + 12

y = + 4

Slope of this line =

Slope of a line perpendicular to the given line =

Required equation of the line passing through A is

y - y1 = m(x - x1)

y - 0 = (x + 3)

4y = -3x - 9

3x + 4y + 9 = 0

Question 22

The point P is the foot of perpendicular from A (-5, 7) to the line whose equation is 2x - 3y + 18 = 0. Determine:

(i) the equation of the line AP

(ii) the co-ordinates of P

Solution 22

(i) The given equation is

2x - 3y + 18 = 0

3y = 2x + 18

y = x + 6

Slope of this line =

Slope of a line perpendicular to this line =

(x1, y1) = (-5, 7)

The required equation of the line AP is given by

y - y1 = m(x - x1)

y - 7 = (x + 5)

2y - 14 = -3x - 15

3x + 2y + 1 = 0

(ii) P is the foot of perpendicular from point A.

So P is the point of intersection of the lines 2x - 3y + 18 = 0 and 3x + 2y + 1 = 0.

2x - 3y + 18 = 0 4x - 6y + 36 = 0

3x + 2y + 1 = 0 9x + 6y + 3 = 0

Adding the two equations, we get,

13x + 39 = 0

x = -3

3y = 2x + 18 = -6 + 18 = 12

y = 4

Thus, the co-ordinates of the point P are (-3, 4).

Question 23

The points A, B and C are (4, 0), (2, 2) and (0, 6) respectively. Find the equations of AB and BC.

If AB cuts the y-axis at P and BC cuts the x-axis at Q, find the co-ordinates of P and Q.

Solution 23

For the line AB:

S l o p e space o f space A B equals m equals fraction numerator 2 minus 0 over denominator 2 minus 4 end fraction equals fraction numerator 2 over denominator minus 2 end fraction equals minus 1

(x1, y1) = (4, 0)

 

Equation of the line AB is

y - y1 = m(x - x1)

y - 0 = -1(x - 4)

y = -x + 4

x + y = 4 ....(1)

 

For the line BC:

S l o p e space o f space B C equals m equals fraction numerator 6 minus 2 over denominator 0 minus 2 end fraction equals fraction numerator 4 over denominator minus 2 end fraction equals minus 2

(x1, y1) = (2, 2)

 

Equation of the line BC is

y - y1 = m(x - x1)

y - 2 = -2(x - 2)

y - 2 = -2x + 4

2x + y = 6 ....(2)

 

Given that AB cuts the y-axis at P. So, the abscissa of point P is 0.

Putting x = 0 in (1), we get,

y = 4

 

Thus, the co-ordinates of point P are (0, 4).

 

Given that BC cuts the x-axis at Q. So, the ordinate of point Q is 0.

Putting y = 0 in (2), we get,

2x = 6 x = 3

 

Thus, the co-ordinates of point Q are (3, 0).

Question 24

Match the equations A, B, C and D with lines L1, L2, L3 and L4, whose graphs are roughly drawn in the given diagram.

A y = 2x; B y - 2x + 2 = 0;

C 3x + 2y = 6; D y = 2

Solution 24

Putting x = 0 and y = 0 in the equation y = 2x, we have:

LHS = 0 and RHS = 0

Thus, the line y = 2x passes through the origin.

Hence, A = L3

Putting x = 0 in y - 2x + 2 = 0, we get, y = -2

Putting y = 0 in y - 2x + 2 = 0, we get, x = 1

So, x-intercept = 1 and y-intercept = -2

So, x-intercept is positive and y-intercept is negative.

Hence, B = L4

Putting x = 0 in 3x + 2y = 6, we get, y = 3

Putting y = 0 in 3x + 2y = 6, we get, x = 2

So, both x-intercept and y-intercept are positive.

Hence, C = L2

The slope of the line y = 2 is 0.

So, the line y = 2 is parallel to x-axis.

Hence, D = L1

Question 25

Find the value of a for which the points A(a, 3), B(2, 1) and C(5, a) are collinear. Hence, find the equation of the line.

Solution 25

Chapter 14 - Equation of a Line Excercise Ex. 14(E)

Question 1

Point P divides the line segment joining the points A (8, 0) and B (16, -8) in the ratio 3: 5. Find its co-ordinates of point P.

Also, find the equation of the line through P and parallel to 3x + 5y = 7.

Solution 1

Using section formula, the co-ordinates of the point P are

 

3x + 5y = 7

Slope of this line =

As the required line is parallel to the line 3x + 5y = 7,

Slope of the required line = Slope of the given line =

Thus, the equation of the required line is

y - y1 = m(x - x1)

y + 3 = (x - 11)

5y + 15 = -3x + 33

3x + 5y = 18

Question 2

The line segment joining the points A(3, -4) and B (-2, 1) is divided in the ratio 1: 3 at point P in it. Find the co-ordinates of P. Also, find the equation of the line through P and perpendicular to the line 5x - 3y + 4 = 0.

Solution 2

Using section formula, the co-ordinates of the point P are

The equation of the given line is

5x - 3y + 4 = 0

Slope of this line =

Since, the required line is perpendicular to the given line,

Slope of the required line =

Thus, the equation of the required line is

y - y1 = m(x - x1)

Question 3

A line 5x + 3y + 15 = 0 meets y-axis at point P. Find the co-ordinates of point P. Find the equation of a line through P and perpendicular to x - 3y + 4 = 0.

Solution 3

Point P lies on y-axis, so putting x = 0 in the equation 5x + 3y + 15 = 0, we get, y = -5

Thus, the co-ordinates of the point P are (0, -5).

x - 3y + 4 = 0

Slope of this line =

The required equation is perpendicular to given equation x - 3y + 4 = 0.

Slope of the required line =

(x1, y1) = (0, -5)

Thus, the required equation of the line is

y - y1 = m(x - x1)

y + 5 = -3(x - 0)

3x + y + 5 = 0

Question 4

Find the value of k for which the lines kx - 5y + 4 = 0 and 5x - 2y + 5 = 0 are perpendicular to each other.

Solution 4

kx - 5y + 4 = 0

Slope of this line = m1 =

5x - 2y + 5 = 0

Slope of this line = m2 =

Since, the lines are perpendicular, m1m2 = -1

Question 5

A straight line passes through the points P (-1, 4) and Q (5, -2). It intersects the co-ordinate axes at points A and B. M is the mid-point of the segment AB. Find:

(i) the equation of the line.

(ii) the co-ordinates of A and B.

(iii) the co-ordinates of M.

Solution 5

(i) Slope of PQ =

Equation of the line PQ is given by

y - y1 = m(x - x1)

y - 4 = -1(x + 1)

y - 4 = -x - 1

x + y = 3

(ii) For point A (on x-axis), y = 0.

Putting y = 0 in the equation of PQ, we get,

x = 3

Thus, the co-ordinates of point A are (3, 0).

For point B (on y-axis), x = 0.

Putting x = 0 in the equation of PQ, we get,

y = 3

Thus, the co-ordinates of point B are (0, 3).

(iii) M is the mid-point of AB.

So, the co-ordinates of point M are

Question 6

(1, 5) and (-3, -1) are the co-ordinates of vertices A and C respectively of rhombus ABCD. Find the equations of the diagonals AC and BD.

Solution 6

A = (1, 5) and C = (-3, -1)

We know that in a rhombus, diagonals bisect each other at right angle.

Let O be the point of intersection of the diagonals AC and BD.

Co-ordinates of O are

Slope of AC =

For line AC:

Slope = m = , (x1, y1) = (1, 5)

Equation of the line AC is

y - y1 = m(x - x1)

y - 5 = (x - 1)

2y - 10 = 3x - 3

3x - 2y + 7 = 0

For line BD:

Slope = m = , (x1, y1) = (-1, 2)

Equation of the line BD is

y - y1 = m(x - x1)

y - 2 = (x + 1)

3y - 6 = -2x - 2

2x + 3y = 4

Question 7

Show that A (3, 2), B (6, -2) and C (2, -5) can be the vertices of a square.

(i) Find the co-ordinates of its fourth vertex D, if ABCD is a square.

(ii) Without using the co-ordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD.

Solution 7

 

U sin g space d i s tan c e space f o r m u l a comma space w e space h a v e :
A B equals square root of open parentheses 6 minus 3 close parentheses squared plus open parentheses minus 2 minus 2 close parentheses squared end root equals square root of 9 plus 16 end root equals 5
B C equals square root of open parentheses 2 minus 6 close parentheses squared plus open parentheses minus 5 plus 2 close parentheses squared end root equals square root of 16 plus 9 end root equals 5
T h u s comma space A C equals B C
A l s o comma space S l o p e space o f space A B equals fraction numerator minus 2 minus 2 over denominator 6 minus 3 end fraction equals fraction numerator minus 4 over denominator 3 end fraction
S l o p e space o f space B C equals fraction numerator minus 5 plus 2 over denominator 2 minus 6 end fraction equals fraction numerator minus 3 over denominator minus 4 end fraction equals 3 over 4
S l o p e space o f space A B cross times S l o p e space o f space B C equals minus 1
T h u s comma space A B perpendicular B C
H e n c e comma space A comma space B comma space C space c a n space b e space t h e space v e r t i c e s space o f space a space s q u a r e..
open parentheses i close parentheses space space S l o p e space o f space A B equals fraction numerator minus 2 minus 2 over denominator 6 minus 3 end fraction equals equals S l o p e space o f space C D
E q u a t i o n space o f space t h e space l i n e space C D space i s
y minus y subscript 1 equals m open parentheses x minus x subscript 1 close parentheses
rightwards double arrow y plus 5 equals fraction numerator minus 4 over denominator 3 end fraction open parentheses x minus 2 close parentheses
rightwards double arrow 3 y plus 15 equals minus 4 x plus 8
rightwards double arrow 4 x plus 3 y equals minus 7.... left parenthesis 1 right parenthesis
S l o p e space o f space B C equals fraction numerator minus 5 plus 2 over denominator 2 minus 6 end fraction equals fraction numerator minus 3 over denominator minus 4 end fraction equals 3 over 4 equals S l o p e space o f space A D
E q u a t i o n space o f space t h e space l i n e space A D space i s
y minus y subscript 1 equals m open parentheses x minus x subscript 1 close parentheses
rightwards double arrow y minus 2 equals 3 over 4 open parentheses x minus 3 close parentheses
rightwards double arrow 4 y minus 8 equals 3 x minus 9
rightwards double arrow 3 x minus 4 y equals 1... left parenthesis 2 right parenthesis
N o w comma space D space i s space t h e space p o i n t space o f space i n t e r s e c t i o n space o f space C D space a n d space A D.
left parenthesis 1 right parenthesis rightwards double arrow 16 x plus 12 y equals minus 28
open parentheses 2 close parentheses rightwards double arrow 9 x minus 12 y equals 3
A d d i n g space t h e space a b o v e space t w o space e q u a t i o n s space w e space g e t comma
25 x equals minus 25
rightwards double arrow x equals minus 1
S o comma space 4 y equals 3 x minus 1 equals minus 3 minus 1 equals minus 4
rightwards double arrow y equals minus 1
T h u s comma space t h e space c o minus o r d i n a t e s space o f space p o i n t space D space a r e space left parenthesis minus 1 comma space minus 1 right parenthesis.

left parenthesis i i right parenthesis
T h e space e q u a t i o n space o f space l i n e space A D space i s space f o u n d space i n space p a r t space left parenthesis i right parenthesis
I t space i s space 3 x minus 4 y equals 1 space o r space 4 y equals 3 x minus 1.
S l o p e space o f space B D equals fraction numerator minus 1 plus 2 over denominator minus 1 minus 6 end fraction equals fraction numerator 1 over denominator minus 7 end fraction equals fraction numerator minus 1 over denominator 7 end fraction
T h e space e q u a t i o n space o f space d i a g o n a l space B D space i s
y minus y subscript 1 equals m open parentheses x minus x subscript 1 close parentheses
rightwards double arrow y plus 1 equals fraction numerator minus 1 over denominator 7 end fraction open parentheses x plus 1 close parentheses
rightwards double arrow 7 y plus 7 equals minus x minus 1
rightwards double arrow x plus 7 y plus 8 equals 0

 

Question 8

A line through origin meets the line x = 3y + 2 at right angles at point X. Find the co-ordinates of X.

Solution 8

The given line is

x = 3y + 2 ...(1)

3y = x - 2

y equals 1 third x minus 2 over 3

Slope of this line is 1 third.

The required line intersects the given line at right angle.

Slope of the required line =fraction numerator minus 1 over denominator begin display style 1 third end style end fraction equals minus 3 

The required line passes through (0, 0) = (x1, y1)

The equation of the required line is

y - y1 = m(x - x1)

y - 0 = -3(x - 0)

3x + y = 0 ...(2)

 

Point X is the intersection of the lines (1) and (2).

Using (1) in (2), we get,

9y + 6 + y = 0

y equals fraction numerator minus 6 over denominator 10 end fraction equals fraction numerator minus 3 over denominator 5 end fraction

Thus, the co-ordinates of the point X are open parentheses 1 fifth comma fraction numerator minus 3 over denominator 5 end fraction close parentheses.

Question 9

A straight line passes through the point (3, 2) and the portion of this line, intercepted between the positive axes, is bisected at this point. Find the equation of the line.

Solution 9

Let the line intersect the x-axis at point A (x, 0) and y-axis at point B (0, y).

Since, P is the mid-point of AB, we have:

Thus, A = (6, 0) and B = (0, 4)

Slope of line AB =

Let (x1, y1) = (6, 0)

The required equation of the line AB is given by

y - y1 = m(x - x1)

y - 0 = (x - 6)

3y = -2x + 12

2x + 3y = 12

Question 10

Find the equation of the line passing through the point of intersection of 7x + 6y = 71 and 5x - 8y = -23; and perpendicular to the line 4x - 2y = 1.

Solution 10

7x + 6y = 71 28x + 24 = 284 ...(1)

5x - 8y = -23 15x - 24y = -69 ...(2)

Adding (1) and (2), we get,

43x = 215

x = 5

 

From (2), 8y = 5x + 23 = 25 + 23 = 48 y = 6

 

Thus, the required line passes through the point (5, 6).

 

4x - 2y = 1

2y = 4x - 1

y = 2x -

Slope of this line = 2

Slope of the required line =

The required equation of the line is

y - y1 = m(x - x1)

y - 6 = (x - 5)

2y - 12 = -x + 5

x + 2y = 17

Question 11

Find the equation of the line which is perpendicular to the line at the point where this line meets y-axis.

Solution 11

The given line is

Slope of this line =

 

Slope of the required line =

 

Let the required line passes through the point P (0, y).

Putting x = 0 in the equation , we get,

Thus, P = (0, -b) = (x1, y1)

 

The equation of the required line is

y - y1 = m(x - x1)

y + b = (x - 0)

by + b2 = -ax

ax + by + b2 = 0

Question 12

O (0, 0), A (3, 5) and B (-5, -3) are the vertices of triangle OAB. Find:

(i) the equation of median of triangle OAB through vertex O.

(ii) the equation of altitude of triangle OAB through vertex B.

Solution 12

(i) Let the median through O meets AB at D. So, D is the mid-point of AB.

Co-ordinates of point D are

Slope of OD =

(x1, y1) = (0, 0)

The equation of the median OD is

y - y1 = m(x - x1)

y - 0 = -1(x - 0)

x + y = 0

(ii) The altitude through vertex B is perpendicular to OA.

Slope of OA =

Slope of the required altitude =

The equation of the required altitude through B is

y - y1 = m(x - x1)

y + 3 = (x + 5)

5y + 15 = -3x - 15

3x + 5y + 30 = 0

Question 13

Determine whether the line through points (-2, 3) and (4, 1) is perpendicular to the line 3x = y + 1.

Does the line 3x = y + 1 bisect the line segment joining the two given points?

Solution 13

Let A = (-2, 3) and B = (4, 1)

Slope of AB = m1 =

Equation of line AB is

y - y1 = m1(x - x1)

y - 3 = (x + 2)

3y - 9 = -x - 2

x + 3y = 7 ...(1)

 

Slope of the given line 3x = y + 1 is 3 = m2.

Hence, the line through points A and B is perpendicular to the given line.

 

Given line is 3x = y +1 ...(2)

 

Solving (1) and (2), we get,

x = 1 and y = 2

 

So, the two lines intersect at point P = (1, 2).

 

The co-ordinates of the mid-point of AB are

Hence, the line 3x = y + 1 bisects the line segment joining the points A and B.

Question 14

Given a straight line x cos + y sin  = 2. Determine the equation of the other line which is parallel to it and passes through (4, 3).

Solution 14

x cos  + y sin  = 2

Slope of this line =

Slope of a line which is parallel to this given line =

Let (4, 3) = (x1, y1)

Thus, the equation of the required line is given by:

y - y1 = m1(x - x1)

y - 3 = (x - 4)

Question 15

Find the value of k such that the line (k - 2)x + (k + 3)y - 5 = 0 is:

(i) perpendicular to the line 2x - y + 7 = 0

(ii) parallel to it.

Solution 15

(k - 2)x + (k + 3)y - 5 = 0 ....(1)

(k + 3)y = -(k - 2)x + 5

y =

Slope of this line = m1 =

(i) 2x - y + 7 = 0

y = 2x + 7 = 0

Slope of this line = m2 = 2

 

Line (1) is perpendicular to 2x - y + 7 = 0

 

(ii) Line (1) is parallel to 2x - y + 7 = 0

Question 16

The vertices of a triangle ABC are A (0, 5), B (-1, -2) and C (11, 7). Write down the equation of BC. Find:

(i) the equation of line through A and perpendicular to BC.

(ii) the co-ordinates of the point, where the perpendicular through A, as obtained in (i), meets BC.

Solution 16

Slope of BC =

Equation of the line BC is given by

y - y1 = m1(x - x1)

y + 2 = (x + 1)

4y + 8 = 3x + 3

3x - 4y = 5....(1)

 

(i) Slope of line perpendicular to BC =

Required equation of the line through A (0, 5) and perpendicular to BC is

y - y1 = m1(x - x1)

y - 5 = (x - 0)

3y - 15 = -4x

4x + 3y = 15 ....(2)

 

(ii) The required point will be the point of intersection of lines (1) and (2).

 

(1) 9x - 12y = 15

(2) 16x + 12y = 60

 

Adding the above two equations, we get,

25x = 75

x = 3

 

So, 4y = 3x - 5 = 9 - 5 = 4

y = 1

 

Thus, the co-ordinates of the required point is (3, 1).

Question 17

From the given figure, find:

(i) the co-ordinates of A, B and C.

(ii) the equation of the line through A and parallel to BC.

Solution 17

(i) A = (2, 3), B = (-1, 2), C = (3, 0)

(ii) Slope of BC =

Slope of required line which is parallel to BC = Slope of BC =

(x1, y1) = (2, 3)

The required equation of the line through A and parallel to BC is given by:

y - y1 = m1(x - x1)

y - 3 = (x - 2)

2y - 6 = -x + 2

x + 2y = 8

Question 18

P (3, 4), Q (7, -2) and R (-2, -1) are the vertices of triangle PQR. Write down the equation of the median of the triangle through R.

Solution 18

The median (say RX) through R will bisect the line PQ.

The co-ordinates of point X are

Slope of RX =

(x1, y1) = (-2, -1)

The required equation of the median RX is given by:

y - y1 = m1(x - x1)

y + 1 = (x + 2)

7y + 7 = 2x + 4

7y = 2x - 3

Question 19

A (8, -6), B (-4, 2) and C (0, -10) are vertices of a triangle ABC. If P is the mid-point of AB and Q is the mid-point of AC, use co-ordinate geometry to show that PQ is parallel to BC. Give a special name of quadrilateral PBCQ.

Solution 19

P is the mid-point of AB. So, the co-ordinate of point P are

Q is the mid-point of AC. So, the co-ordinate of point Q are

Slope of PQ =

Slope of BC =

Since, slope of PQ = Slope of BC,

PQ || BC

Also, we have:

Slope of PB =

Slope of QC =

Thus, PB is not parallel to QC.

Hence, PBCQ is a trapezium.

Question 20

A line AB meets the x-axis at point A and y-axis at point B. The point P (-4, -2) divides the line segment AB internally such that AP : PB = 1 : 2. Find:

(i) the co-ordinates of A and B.

(ii) the equation of line through P and perpendicular to AB.

Solution 20

(i) Let the co-ordinates of point A (lying on x-axis) be (x, 0) and the co-ordinates of point B (lying y-axis) be (0, y).

Given, P = (-4, -2) and AP: PB = 1:2

Using section formula, we have:

Thus, the co-ordinates of A and B are (-6, 0) and (0, -6).

(ii) Slope of AB =

Slope of the required line perpendicular to AB =

(x1, y1) = (-4, -2)

Required equation of the line passing through P and perpendicular to AB is given by

y - y1 = m(x - x1)

y + 2 = 1(x + 4)

y + 2 = x + 4

y = x + 2

Question 21

A line intersects x-axis at point (-2, 0) and cuts off an intercept of 3 units from the positive side of y-axis. Find the equation of the line.

Solution 21

The required line intersects x-axis at point A (-2, 0).

Also, y-intercept = 3

So, the line also passes through B (0, 3).

Slope of line AB = = m

(x1, y1) = (-2, 0)

Required equation of the line AB is given by

y - y1 = m(x - x1)

y - 0 = (x + 2)

2y = 3x + 6

Question 22

Find the equation of a line passing through the point (2, 3) and having the x-intercept of 4 units.

Solution 22

The required line passes through A (2, 3).

Also, x-intercept = 4

So, the required line passes through B (4, 0).

Slope of AB =

(x1, y1) = (4, 0)

Required equation of the line AB is given by

y - y1 = m(x - x1)

y - 0 = (x - 4)

2y = -3x + 12

3x + 2y = 12

Question 23

The given figure (not drawn to scale) shows two straight lines AB and CD. If equation of the line AB is: y = x + 1 and equation of line CD is: y = x - 1. Write down the inclination of lines AB and CD; also, find the angle between AB and CD.

 

Solution 23

Equation of the line AB is y = x + 1

Slope of AB = 1

Inclination of line AB = (Since, tan 45o = 1)

Equation of line CD is y = x - 1

Slope of CD =

Inclination of line CD = 60o (Since, tan 60o =)

Using angle sum property in PQR,

Question 24

Write down the equation of the line whose gradient is and which passes through P, where P divides the line segment joining A (-2, 6) and B (3, -4) in the ratio 2: 3.

Solution 24

Given, P divides the line segment joining A (-2, 6) and B (3, -4) in the ratio 2: 3.

Co-ordinates of point P are

Slope of the required line = m =

The required equation of the line is given by

y - y1 = m(x - x1)

y - 2 = (x - 0)

2y - 4 = 3x

2y = 3x + 4

Question 25

The ordinate of a point lying on the line joining the points (6, 4) and (7, -5) is -23. Find the co-ordinates of that point.

Solution 25

Let A = (6, 4) and B = (7, -5)

Slope of the line AB =

(x1, y1) = (6, 4)

 

The equation of the line AB is given by

y - y1 = m(x - x1)

y - 4 = -9(x - 6)

y - 4 = -9x + 54

9x + y = 58 ...(1)

 

Now, given that the ordinate of the required point is -23.

Putting y = -23 in (1), we get,

9x - 23 = 58

9x = 81

x = 9

 

Thus, the co-ordinates of the required point is (9, -23).

Question 26

Points A and B have coordinates (7, -3) and (1, 9) respectively. Find:

(i) the slope of AB.

(ii) the equation of the perpendicular bisector of the line segment AB.

(iii) the value of 'p' if (-2, p) lies on it.

Solution 26

Given points are A(7, -3) and B(1, 9).

(i) Slope of AB =

(ii) Slope of perpendicular bisector = =

Mid-point of AB = =(4, 3)

Equation of perpendicular bisector is:

y - 3 = (x - 4)

2y - 6 = x - 4

x - 2y + 2 = 0

(iii) Point (-2, p) lies on x - 2y + 2 = 0.

-2 - 2p + 2 = 0

2p = 0

p = 0

Question 27

A and B are two points on the x-axis and y-axis respectively. P (2, -3) is the mid-point of AB. Find the

(i) coordinates of A and B

(ii) slope of line AB

(iii) equation of line AB.

Solution 27

(i) Let the co-ordinates be A(x, 0) and B(0, y).

Mid-point of A and B is given by

(ii) Slope of line AB, m =

(iii) Equation of line AB, using A(4, 0)

2y = 3x - 12

Question 28

The equation of a line 3x + 4y - 7 = 0. Find:

(i) the slope of the line.

(ii) the equation of a line perpendicular to the given line and passing through the intersection of the lines x - y + 2 = 0 and 3x + y - 10 = 0.

Solution 28

3x + 4y - 7 = 0 ...(1)

4y = -3x + 7

y =

(i) Slope of the line = m =

(ii) Slope of the line perpendicular to the given line =

Solving the equations x - y + 2 = 0 and 3x + y - 10 = 0, we get x = 2 and y = 4.

So, the point of intersection of the two given lines is (2, 4).

Given that a line with slope passes through point (2, 4).

Thus, the required equation of the line is

y - 4 = (x - 2)

3y - 12 = 4x - 8

4x - 3y + 4 = 0

Question 29

ABCD is a parallelogram where A(x, y), B(5, 8), C(4, 7) and D(2, -4). Find:

(i)Co-ordinates of A

(ii)Equation of diagonal BD

Solution 29

 

In parallelogram ABCD, A(x, y), B(5, 8), C(4, 7) and D(2, -4).

The diagonals of the parallelogram bisect each other.

O is the point of intersection of AC and BD

Since O is the midpoint of BD, its coordinates will be

 

(i)

Since O is the midpoint of AC also,

(ii)

Question 30

Given equation of the line L1 is y = 4.

(i)Write the slope of the line L2 if L2 is the bisector of angle O

(ii)Write the coordinates of point P

(iii)Find the equation of L2


Solution 30

(i)

 

(ii)

 

 

(iii)

Question 31

(i) equation of AB

(ii) equation of CD

 

Solution 31

Question 32

Find the equation of the line that has x-intercept = -3 and is perpendicular to 3x + 5y = 1.

Solution 32

Question 33

A straight line passes through the points P(-1, 4) and Q(5, -2). It intersects x-axis at point A and y-axis at point B. M is the mid- t point of the line segment AB. Find:

(i) the equation of the line.

(ii) the co-ordinates of points A and B.

(iii) the co-ordinates of point M

Solution 33

Question 34

In the given figure. line AB meets y-axis at point A. Line through C(2, 10) and D intersects line AB at right angle at point R Find:

(i) equation of line AB

(ii) equation of line CD

(iii) co-ordinates of points E and D

 

Solution 34

Question 35

A line through point P(4, 3) meets x-axis at point A and the y-axis at point B. If BP is double of PA, find the equation of AB.

Solution 35

Question 36

Find the equation of line through the intersection of lines 2x - y = 1 and 3x + 2y = -9 and making an angle of 30° with positive direction of x-axis.

Solution 36

Question 37

Find the equation of the line through the Points A(-1, 3) and B(0, 2). Hence, show that the points A, B and C(1, 1) are collinear.

Solution 37

Question 38

Three vertices of a parallelogram ABCD taken in order are A(3, 6), B(5, 10) and C(3, 2), find :

(i) the co-ordinates of the fourth vertex D.

(ii) length of diagonal BD.

(iii) equation of side AB of the parallelogram ABCD.

Solution 38

Question 39

In the figure, given, ABC is a triangle and BC is parallel to the y-axis. AB and AC intersect the y-axis at P and Q respectively.

(i) Write the co-ordinates of A.

(ii) Find the length of AB and AC.

(iii) Find the radio in which Q divides AC.

(iv) Find the equation of the line AC.

Solution 39

Question 40

 i. k.

 ii. mid-point of PQ, using the value of 'k' found in (i). 

Solution 40

  

 

 

Question 41

A line AB meets X-axis at A and Y-axis at B. P(4, -1) divides AB in the ration 1 : 2.

 i. Find the co-ordinates of A and B.

 ii. Find the equation of the line through P and perpendicular to AB.

 

  

 

Solution 41

i. Since A lies on the X-axis, let the co-ordinates of A be (x, 0).

Since B lies on the Y-axis, let the co-ordinates of B be (0, y).

Let m = 1 and n = 2

Using Section formula,

x = 6 and y = -3

So, the co-ordinates of A are (6, 0) and that of B are (0, -3). 

 

Slope of line perpendicular to AB = m = -2

P = (4, -1)

Thus, the required equation is

y - y1 = m(x - x1)

y - (-1) = -2(x - 4)

y + 1 = -2x + 8

2x + y = 7 

Chapter 14 - Equation of a Line Excercise Ex. 14(A)

Question 1

Find, which of the following points lie on the line x - 2y + 5 = 0:

(i) (1, 3) (ii) (0, 5)

(iii) (-5, 0) (iv) (5, 5)

(v) (2, -1.5) (vi) (-2, -1.5)

Solution 1

The given line is x - 2y + 5 = 0.


(i) Substituting x = 1 and y = 3 in the given equation, we have:

L.H.S. = 1 - 2 3 + 5 = 1 - 6 + 5 = 6 - 6 = 0 = R.H.S.

Thus, the point (1, 3) lies on the given line.


 

(ii) Substituting x = 0 and y = 5 in the given equation, we have:

L.H.S. = 0 - 2 5 + 5 = -10 + 5 = -5 R.H.S.

Thus, the point (0, 5) does not lie on the given line.


 

(iii) Substituting x = -5 and y = 0 in the given equation, we have:

L.H.S. = -5 - 2 0 + 5 = -5 - 0 + 5 = 5 - 5 = 0 = R.H.S.

Thus, the point (-5, 0) lie on the given line.


 

(iv) Substituting x = 5 and y = 5 in the given equation, we have:

L.H.S. = 5 - 2 5 + 5 = 5 - 10 + 5 = 10 - 10 = 0 = R.H.S.

Thus, the point (5, 5) lies on the given line.


 

(v) Substituting x = 2 and y = -1.5 in the given equation, we have:

L.H.S. = 2 - 2 (-1.5) + 5 = 2 + 3 + 5 = 10 R.H.S.

Thus, the point (2, -1.5) does not lie on the given line.


 

(vi) Substituting x = -2 and y = -1.5 in the given equation, we have:

L.H.S. = -2 - 2 (-1.5) + 5 = -2 + 3 + 5 = 6 R.H.S.

Thus, the point (-2, -1.5) does not lie on the given line.

Question 2

State, true or false:

(i) the line passes through the point (2, 3).

(ii) the line passes through the point (4, -6).

(iii) the point (8, 7) lies on the line y - 7 = 0.

(iv) the point (-3, 0) lies on the line x + 3 = 0.

(v) if the point (2, a) lies on the line 2x - y = 3, then a = 5.

Solution 2

(i) The given line is

Substituting x = 2 and y = 3 in the given equation,

Thus, the given statement is false.

(ii) The given line is

Substituting x = 4 and y = -6 in the given equation,

Thus, the given statement is true.

(iii) L.H.S = y - 7 = 7 - 7 = 0 = R.H.S.

Thus, the point (8, 7) lies on the line y - 7 = 0.

The given statement is true.

(iv) L.H.S. = x + 3 = -3 + 3 = 0 = R.H.S

Thus, the point (-3, 0) lies on the line x + 3 = 0.

The given statement is true.

(v) The point (2, a) lies on the line 2x - y = 3.

2(2) - a = 3

4 - a = 3

a = 4 - 3 = 1

Thus, the given statement is false.

Question 3

The line given by the equation passes through the point (k, 6); calculate the value of k.

Solution 3

Given, the line given by the equation passes through the point (k, 6).

Substituting x = k and y = 6 in the given equation, we have:

Question 4

For what value of k will the point (3, -k) lie on the line 9x + 4y = 3?

Solution 4

The given equation of the line is 9x + 4y = 3.

Put x = 3 and y = -k, we have:

9(3) + 4(-k) = 3

27 - 4k = 3

4k = 27 - 3 = 24

k = 6

Question 5

The line contains the point (m, 2m - 1); calculate the value of m.

Solution 5

The equation of the given line is

Putting x = m, y = 2m - 1, we have:


Question 6

Does the line 3x - 5y = 6 bisect the join of (5, -2) and (-1, 2)?

Solution 6

The given line will bisect the join of A (5, -2) and B (-1, 2), if the co-ordinates of the mid-point of AB satisfy the equation of the line.

The co-ordinates of the mid-point of AB are

Substituting x = 2 and y = 0 in the given equation, we have:

L.H.S. = 3x - 5y = 3(2) - 5(0) = 6 - 0 = 6 = R.H.S.

Hence, the line 3x - 5y = 6 bisect the join of (5, -2) and (-1, 2).

Question 7

(i) The line y = 3x - 2 bisects the join of (a, 3) and (2, -5), find the value of a.

(ii) The line x - 6y + 11 = 0 bisects the join of (8, -1) and (0, k). Find the value of k.

Solution 7

(i) The given line bisects the join of A (a, 3) and B (2, -5), so the co-ordinates of the mid-point of AB will satisfy the equation of the line.

The co-ordinates of the mid-point of AB are

Substituting x = and y = -1 in the given equation, we have:

(ii) The given line bisects the join of A (8, -1) and B (0, k), so the co-ordinates of the mid-point of AB will satisfy the equation of the line.

The co-ordinates of the mid-point of AB are

Substituting x = 4 and y = in the given equation, we have:

Question 8

(i) The point (-3, 2) lies on the line ax + 3y + 6 = 0, calculate the value of a.

(ii) The line y = mx + 8 contains the point (-4, 4), calculate the value of m.

Solution 8

(i) Given, the point (-3, 2) lies on the line ax + 3y + 6 = 0.

Substituting x = -3 and y = 2 in the given equation, we have:

a(-3) + 3(2) + 6 = 0

-3a + 12 = 0

3a = 12

a = 4

(ii) Given, the line y = mx + 8 contains the point (-4, 4).

Substituting x = -4 and y = 4 in the given equation, we have:

4 = -4m + 8

4m = 4

m = 1

Question 9

The point P divides the join of (2, 1) and (-3, 6) in the ratio 2: 3. Does P lie on the line x - 5y + 15 = 0?

Solution 9

Given, the point P divides the join of (2, 1) and (-3, 6) in the ratio 2: 3.

Co-ordinates of the point P are

Substituting x = 0 and y = 3 in the given equation, we have:

L.H.S. = 0 - 5(3) + 15 = -15 + 15 = 0 = R.H.S.

Hence, the point P lies on the line x - 5y + 15 = 0.

Question 10

The line segment joining the points (5, -4) and (2, 2) is divided by the point Q in the ratio 1: 2. Does the line x - 2y = 0 contain Q?

Solution 10

Given, the line segment joining the points (5, -4) and (2, 2) is divided by the point Q in the ratio 1: 2.

Co-ordinates of the point Q are

Substituting x = 4 and y = -2 in the given equation, we have:

L.H.S. = x - 2y = 4 - 2(-2) = 4 + 4 = 8 R.H.S.

Hence, the given line does not contain point Q.

Question 11

Find the point of intersection of the lines:

4x + 3y = 1 and 3x - y + 9 = 0. If this point lies on the line (2k - 1)x - 2y = 4; find the value of k.

Solution 11


Consider the given equations:

4x + 3y = 1 ....(1)

3x - y + 9 = 0 ....(2)

Multiplying (2) with 3, we have:

9x - 3y = -27 ....(3)

Adding (1) and (3), we get,

13x = -26

x = -2

 

From (2), y = 3x + 9 = -6 + 9 = 3

 

Thus, the point of intersection of the given lines (1) and (2) is (-2, 3).

 

The point (-2, 3) lies on the line (2k - 1)x - 2y = 4.

(2k - 1)(-2) - 2(3) = 4

-4k + 2 - 6 = 4

-4k = 8

k = -2

Question 12

Show that the lines 2x + 5y = 1, x - 3y = 6 and x + 5y + 2 = 0 are concurrent.

Solution 12

We know that two or more lines are said to be concurrent if they intersect at a single point.

 

We first find the point of intersection of the first two lines.

2x + 5y = 1 ....(1)

x - 3y = 6 ....(2)

 

Multiplying (2) by 2, we get,

2x - 6y = 12 ....(3)

 

Subtracting (3) from (1), we get,

11y = -11

y = -1

 

From (2), x = 6 + 3y = 6 - 3 = 3

 

So, the point of intersection of the first two lines is (3, -1).

If this point lie on the third line, i.e., x + 5y + 2 = 0, then the given lines will be concurrent.

 

Substituting x = 3 and y = -1, we have:

L.H.S. = x + 5y + 2 = 3 + 5(-1) + 2 = 5 - 5 = 0 = R.H.S.

 

Thus, (3, -1) also lie on the third line.

Hence, the given lines are concurrent.

Chapter 14 - Equation of a Line Excercise Ex. 14(B)

Question 1

Find the slope of the line whose inclination is:

(i) 0 (ii) 30

(iii) 72 30' (iv) 46

Solution 1

(i) Slope = tan 0o = 0

(ii) Slope = tan 30o =

(iii) Slope = tan 72o 30' = 3.1716

(iv) Slope = tan 46o = 1.0355

Question 2

Find the inclination of the line whose slope is:

(i) 0 (ii)

(iii) 0.7646 (iv) 1.0875

Solution 2

(i) Slope = tan = 0

= 0o

(ii) Slope = tan =

= 60o

(iii) Slope = tan = 0.7646

= 37o 24'

(iv) Slope = tan = 1.0875

= 47o 24'

Question 3

Find the slope of the line passing through the following pairs of points:

(i) (-2, -3) and (1, 2)

(ii) (-4, 0) and origin

(iii) (a, -b) and (b, -a)

Solution 3

We know:

Slope =

(i) Slope =

(ii) Slope =

(iii) Slope =

Question 4

Find the slope of the line parallel to AB if:

(i) A = (-2, 4) and B = (0, 6)

(ii) A = (0, -3) and B = (-2, 5)

Solution 4

(i) Slope of AB =

Slope of the line parallel to AB = Slope of AB = 1

(ii) Slope of AB =

Slope of the line parallel to AB = Slope of AB = -4

Question 5

Find the slope of the line perpendicular to AB if:

(i) A = (0, -5) and B = (-2, 4)

(ii) A = (3, -2) and B = (-1, 2)

Solution 5

(i) Slope of AB =

Slope of the line perpendicular to AB =

(ii) Slope of AB =

Slope of the line perpendicular to AB = 1

Question 6

The line passing through (0, 2) and (-3, -1) is parallel to the line passing through (-1, 5) and (4, a). Find a.

Solution 6

Slope of the line passing through (0, 2) and (-3, -1) =

Slope of the line passing through (-1, 5) and (4, a) =

Since, the lines are parallel.

Question 7

The line passing through (-4, -2) and (2, -3) is perpendicular to the line passing through (a, 5) and (2, -1). Find a.

Solution 7

Slope of the line passing through (-4, -2) and (2, -3) =

Slope of the line passing through (a, 5) and (2, -1) =

Since, the lines are perpendicular.

Question 8

Without using the distance formula, show that the points A (4, -2), B (-4, 4) and C (10, 6) are the vertices of a right-angled triangle.

Solution 8

The given points are A (4, -2), B (-4, 4) and C (10, 6).


 S l o p e space o f space A B equals fraction numerator 4 plus 2 over denominator minus 4 minus 4 end fraction equals fraction numerator 6 over denominator minus 8 end fraction equals fraction numerator minus 3 over denominator 4 end fraction

S l o p e space o f space B C equals fraction numerator 6 minus 4 over denominator 10 plus 4 end fraction equals 2 over 14 equals 1 over 7

S l o p e space o f space A C equals fraction numerator 6 plus 2 over denominator 10 minus 4 end fraction equals 8 over 6 equals 4 over 3

It can be seen that:

S l o p e space o f space A B equals fraction numerator minus 1 over denominator S l o p e space o f space A C end fraction


Hence, AB AC.


Thus, the given points are the vertices of a right-angled triangle.

Question 9

Without using the distance formula, show that the points A (4, 5), B (1, 2), C (4, 3) and D (7, 6) are the vertices of a parallelogram.

Solution 9

The given points are A (4, 5), B (1, 2), C (4, 3) and D (7, 6).


S l o p e space o f space A B equals fraction numerator 2 minus 5 over denominator 1 minus 4 end fraction equals fraction numerator minus 3 over denominator minus 3 end fraction equals 1

S l o p e space o f space C D equals fraction numerator 6 minus 3 over denominator 7 minus 4 end fraction equals 3 over 3 equals 1


Since, slope of AB = slope of CD

 

Therefore AB || CD

 

S l o p e space o f space B C equals fraction numerator 3 minus 2 over denominator 4 minus 1 end fraction equals 1 third
S l o p e space o f space D A equals fraction numerator 5 minus 6 over denominator 4 minus 7 end fraction equals fraction numerator minus 1 over denominator minus 3 end fraction equals 1 third

 

 

Since, slope of BC = slope of DA


Therefore, BC || DA

 

Hence, ABCD is a parallelogram

Question 10

(-2, 4), (4, 8), (10, 7) and (11, -5) are the vertices of a quadrilateral. Show that the quadrilateral, obtained on joining the mid-points of its sides, is a parallelogram.

Solution 10

Let the given points be A (-2, 4), B (4, 8), C (10, 7) and D (11, -5).


Let P, Q, R and S be the mid-points of AB, BC, CD and DA respectively.

 

Co-ordinates of P are



 

Co-ordinates of Q are



 

Co-ordinates of R are



 

Co-ordinates of S are



S l o p e space o f space P Q equals fraction numerator begin display style 15 over 2 end style minus 6 over denominator 7 minus 1 end fraction equals fraction numerator begin display style fraction numerator 15 minus 12 over denominator 2 end fraction end style over denominator 6 end fraction equals 3 over 12 equals 1 fourth
S l o p e space o f space R S equals fraction numerator begin display style fraction numerator minus 1 over denominator 2 end fraction end style minus 1 over denominator begin display style 9 over 2 end style minus begin display style 21 over 2 end style end fraction equals fraction numerator begin display style fraction numerator minus 1 minus 2 over denominator 2 end fraction end style over denominator begin display style fraction numerator 9 minus 21 over denominator 2 end fraction end style end fraction equals fraction numerator minus 3 over denominator minus 12 end fraction equals 1 fourth


 

 

Since, slope of PQ = Slope of RS, PQ || RS.


S l o p e space o f space Q R equals fraction numerator 1 minus begin display style 15 over 2 end style over denominator begin display style 21 over 2 end style minus 7 end fraction equals fraction numerator begin display style fraction numerator 2 minus 15 over denominator 2 end fraction end style over denominator begin display style fraction numerator 21 minus 14 over denominator 2 end fraction end style end fraction equals fraction numerator minus 13 over denominator 7 end fraction
S l o p e space o f space S P equals fraction numerator 6 plus begin display style 1 half end style over denominator 1 minus begin display style 9 over 2 end style end fraction equals fraction numerator begin display style fraction numerator 12 plus 1 over denominator 2 end fraction end style over denominator begin display style fraction numerator 2 minus 9 over denominator 2 end fraction end style end fraction equals fraction numerator 13 over denominator minus 7 end fraction equals fraction numerator minus 13 over denominator 7 end fraction

 

 

 

Since, slope of QR = Slope of SP, QR || SP.


Hence, PQRS is a parallelogram.

Question 11

Show that the points P (a, b + c), Q (b, c + a) and R (c, a + b) are collinear.

Solution 11

The points P, Q, R will be collinear if slope of PQ and QR is the same.

 

 

S l o p e space o f space P Q equals fraction numerator c plus a minus b minus c over denominator b minus a end fraction equals fraction numerator a minus b over denominator b minus a end fraction equals minus 1
S l o p e space o f space Q R equals fraction numerator a plus b minus c minus a over denominator c minus b end fraction equals fraction numerator b minus c over denominator c minus b end fraction equals minus 1


 

Hence, the points P, Q, and R are collinear.

Question 12

Find x, if the slope of the line joining (x, 2) and (8, -11) is.

Solution 12

Let A = (x, 2) and B = (8, -11)

Slope of AB =

Question 13

The side AB of an equilateral triangle ABC is parallel to the x-axis. Find the slope of all its sides.

Solution 13

We know that the slope of any line parallel to x-axis is 0.

Therefore, slope of AB = 0

 

Since, ABC is an equilateral triangle, 

Slope of AC = tan 60o =

 

Slope of BC = -tan 60o = -

Question 14

The side AB of a square ABCD is parallel to the x-axis. Find the slopes of all its sides.

Also, find:

(i) the slope of the diagonal AC,

(ii) the slope of the diagonal BD.

 

Solution 14

We know that the slope of any line parallel to x-axis is 0.

Therefore, slope of AB = 0

 

As CD || BC, slope of CD = Slope of AB = 0

 

As BC AB, slope of BC =

As AD AB, slope of AD =

 

(i) The diagonal AC makes an angle of 45o with the positive direction of x axis.

(ii) The diagonal BC makes an angle of -45o with the positive direction of x axis.

Question 15

A (5, 4), B (-3, -2) and C (1, -8) are the vertices of a triangle ABC. Find:

(i) the slope of the altitude of AB,

(ii) the slope of the median AD, and

(iii) the slope of the line parallel to AC.

Solution 15

Given, A (5, 4), B (-3, -2) and C (1, -8) are the vertices of a triangle ABC.

 

(i) Slope of AB =

Slope of the altitude of AB =

 

(ii) Since, D is the mid-point of BC.

Co-ordinates of point D are

Slope of AD =

 

(iii) Slope of AC =

Slope of line parallel to AC = Slope of AC = 3

Question 16

The slope of the side BC of a rectangle ABCD is. Find:

(i) the slope of the side AB,

(ii) the slope of the side AD.

Solution 16

(i) Since, BC is perpendicular to AB,

Slope of AB =

 

(ii) Since, AD is parallel to BC,

Slope of AD = Slope of BC =

Question 17

Find the slope and the inclination of the line AB if:

(i) A = (-3, -2) and B = (1, 2)

(ii) A = (0, ) and B = (3, 0)

(iii) A = (-1, 2) and B = (-2, )

Solution 17

(i) A = (-3, -2) and B = (1, 2)

Slope of AB =

Inclination of line AB = = 45o

 

(ii) A = (0, ) and B = (3, 0)

Slope of AB =

Inclination of line AB = = 30o

 

(iii) A = (-1, 2) and B = (-2, )

Slope of AB =

Inclination of line AB = = 60o

Question 18

The points (-3, 2), (2, -1) and (a, 4) are collinear. Find a.

Solution 18

Given, points A (-3, 2), B (2, -1) and C (a, 4) are collinear.

Slope of AB = Slope of BC

Question 19

The points (K, 3), (2, -4) and (-K + 1, -2) are collinear. Find K.

Solution 19

Given, points A (K, 3), B (2, -4) and C (-K + 1, -2) are collinear.

Slope of AB = Slope of BC

Question 20

Plot the points A (1, 1), B (4, 7) and C (4, 10) on a graph paper. Connect A and B, and also A and C.

Which segment appears to have the steeper slope, AB or AC?

Justify your conclusion by calculating the slopes of AB and AC.

Solution 20

 

From the graph, clearly, AC has steeper slope.

 

Slope of AB =

Slope of AC =

The line with greater slope is steeper. Hence, AC has steeper slope.

Question 21

Find the value(s) of k so that PQ will be parallel to RS. Given:

(i) P (2, 4), Q (3, 6), R (8, 1) and S (10, k)

(ii) P (3, -1), Q (7, 11), R (-1, -1) and S (1, k)

(iii) P (5, -1), Q (6, 11), R (6, -4k) and S (7, k2)

Solution 21

Since, PQ || RS,

Slope of PQ = Slope of RS

 

(i) Slope of PQ =

Slope of RS =

 

(ii) Slope of PQ =

Slope of RS =

 

(iii) Slope of PQ =

Slope of RS =

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