# SELINA Solutions for Class 10 Maths Chapter 14 - Equation of a Line

Study well with our online Selina Solutions for ICSE Class 10 Mathematics Chapter 14 Equation of a Line. Learn to work out the slope of a given line which is parallel or perpendicular to another given line. Also, grasp the concept of slope-intercept form and two-point form to find answers for problems based on equation of a line.

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

Find the equation of a line whose:

y-intercept = 2 and slope = 3.

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.

Find the equation of a line whose:

y-intercept
= -1 and inclination = 45^{o}.

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

Slope = m = tan 45^{o}
= 1

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

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

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

Given, slope =

The equation passes
through (-3, 4) = (x_{1}, y_{1})

Substituting the
values in y - y_{1} = m(x - x_{1}), we get,

y - 4 = (x + 3)

3y - 12 = -4x - 12

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

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

Slope of the line =
tan 60^{o} =

The line passes
through the point (5, 4) = (x_{1}, y_{1})

Substituting the
values in y - y_{1} = m(x - x_{1}), we get,

y - 4 = (x - 5)

y - 4 = x - 5

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

Find the equation of the line passing through:

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

(i) Let (0, 1) = (x_{1},
y_{1}) and (1, 2) = (x_{2}, y_{2})

The required equation of the line is given by:

y - y_{1} =
m(x - x_{1})

y - 1 = 1(x - 0)

y - 1 = x

y = x + 1

(ii) Let (-1, -4) =
(x_{1}, y_{1}) and (3, 0) = (x_{2}, y_{2})

The required equation of the line is given by:

y - y_{1} =
m(x - x_{1})

y + 4 = 1(x + 1)

y + 4 = x + 1

y = x - 3

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.

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 - y_{1} =
m(x - x_{1})

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).

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.

(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 - y_{1} =
m(x - x_{1})

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).

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.

Slope of line AB =
tan 45^{o} = 1

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

y - y_{1} =
m(x - x_{1})

y - 4 = 1(x - 3)

y - 4 = x - 3

y = x + 1

Slope of line CD =
tan 60^{o} =

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

y - y_{1} =
m(x - x_{1})

y - 4 = (x - 3)

y - 4 = x - 3

y = x + 4 - 3

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

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

Since, ABCD is a parallelogram,

B = 180^{o}
- 60^{o} = 120^{o}

Slope
of BC = tan 120^{o} = tan (90^{o} + 30^{o}) = cot30^{o}
=

Equation of the line BC is given by:

y - y_{1}
= m(x - x_{1})

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 - y_{1}
= m(x - x_{1})

y - 5 = 0(x - 7)

y = 5

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.

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).

Required equation of the line is given by:

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.

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 - y_{1} =
m(x - x_{1})

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

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.

Slope of BC =

Slope of line perpendicular to BC =

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

y - y_{1} =
m(x - x_{1})

y - 3 = 1(x - 0)

y - 3 = x

y = x + 3

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

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 - y_{1} =
m(x - x_{1})

y - 2 = 1(x + 1)

y - 2 = x + 1

y = x + 3

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

(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 (x_{1}, y_{1}) = (5, 0) and (x_{2}, y_{2}) = (0, 3)

Slope =

The required equation is:

y - y_{1} = m(x - x_{1})

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 (x_{1}, y_{1}) = (-4, 0) and (x_{2}, y_{2}) = (0, 6)

Slope =

The required equation is:

y - y_{1} = m(x - x_{1})

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 (x_{1}, y_{1}) = (-8, 0) and (x_{2}, y_{2}) = (0, -4)

Slope =

The required equation is:

y - y_{1} = m(x - x_{1})

y - 0 = (x + 8)

2y = -x - 8

x + 2y + 8 = 0

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

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 - y_{1} =
m(x - x_{1})

y - 0 = (x - 6)

6y = -5x + 30

5x + 6y = 30

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

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 - y_{1} =
m(x - x_{1})

y - 0 = (x - 5)

4y = -x + 5

x + 4y = 5

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

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 - y_{1} =
m(x - x_{1})

y - 5 = -2(x - 0)

y - 5 = -2x

2x + y = 5

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

Let AB and CD be two equally inclined lines.

**For line AB:**

Slope = m = tan 45^{o}
= 1

(x_{1}, y_{1})
= (-2, 0)

Equation of the line AB is:

y - y_{1} =
m(x - x_{1})

y - 0 = 1(x + 2)

y = x + 2

**For line CD:**

Slope = m = tan
(-45^{o}) = -1

(x_{1}, y_{1})
= (-2, 0)

Equation of the line CD is:

y - y_{1} =
m(x - x_{1})

y - 0 = -1(x + 2)

y = -x - 2

x + y + 2 = 0

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.

(i)

(ii)

(iii)

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.

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 - y_{1} =
m(x - x_{1})

y + 2 = (x - 10)

5y + 10 = -2x + 20

2x + 5y = 10

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.

(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 - y_{1} = m(x - x_{1})

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.

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 - y_{1} =
m(x - x_{1})

y - 1 = 0(x - 3)

y = 1

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

Find the slope and y-intercept of the line:

(i) y = 4

(ii) ax - by = 0

(iii) 3x - 4y = 5

(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 =

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

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 45^{o}

(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.

(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.

Find the slope of the line which is parallel to:

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

(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 =

Find the slope of the line which is perpendicular to:

(i) (ii)

(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 =

(i) Lines 2x - by + 5 = 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.

(i) 2x - by + 5 = 0

by = 2x + 5

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.

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

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.

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.

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

2y = 2x + 3

y = x +

Slope of the line AB = 1

(ii) Required angle =

Slope = tan = 1 = tan 45^{o}

= 45^{o}

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

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.

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

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.

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

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

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

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

(x_{1}, y_{1}) = (-5, 7)

Required equation of the line is

y - y_{1} = m(x - x_{1})

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.

(x_{1}, y_{1}) = (-5, 7)

Required equation of the line is

x - x_{1 }=0

x + 5=0

(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).

(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 - y_{1} =
m(x - x_{1})

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 - y_{1 = } m(x
- x_{1})

y - 1 =

2y - 2 = -3x

3x + 2y = 2

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

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 - y_{1} =
m(x - x_{1})

y - 1 = (x + 2)

4y - 4 = 5x + 10

5x - 4y + 14 = 0

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

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 - y_{1} =
m(x - x_{1})

y - 0 = 1(x - 3)

y = x - 3

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.

(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 - y_{1} =
m(x - x_{1})

y - 3 = (x - 2)

2y - 6 = -x + 2

X + 2y = 8

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

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 = , (x_{1}, y_{1})
= (-5, 6)

Equation of the line BD is

y - y_{1} =
m(x - x_{1})

y - 6 = (x + 5)

3y - 18 = -x - 5

x + 3y = 13

For line AC:

Slope = m = , (x_{1}, y_{1})
= (-2, 5)

Equation of the line AC is

y - y_{1} =
m(x - x_{1})

y - 5 = 3(x + 2)

y - 5 = 3x + 6

y = 3x + 11

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

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 = , (x_{1}, y_{1})
= (7, -2)

Equation of the line AC is

y - y_{1} =
m(x - x_{1})

y + 2 = (x - 7)

2y + 4 = x - 7

2y = x - 11

For line BD:

Slope = m = , (x_{1}, y_{1})
= (3, -4)

Equation of the line BD is

y - y_{1} =
m(x - x_{1})

y + 4 = -2(x - 3)

y + 4 = -2x + 6

2x + y = 2

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.

(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

(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.

(i) 2y = 3x + 5

Slope of this line =

Slope of the line AB =

(x_{1}, y_{1})
= (3, 2)

The required equation of the line AB is

y - y_{1} =
m(x - x_{1})

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

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.

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, (x_{1},
y_{1}) = (-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 - y_{1} =
m(x - x_{1})

y - 0 = (x + 3)

4y = -3x - 9

3x + 4y + 9 = 0

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

(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 =

(x_{1}, y_{1})
= (-5, 7)

The required equation of the line AP is given by

y - y_{1} =
m(x - x_{1})

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).

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.

For the line AB:

(x_{1}, y_{1}) = (4, 0)

Equation of the line AB is

y - y_{1} = m(x - x_{1})

y - 0 = -1(x - 4)

y = -x + 4

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

For the line BC:

(x_{1}, y_{1}) = (2, 2)

Equation of the line BC is

y - y_{1} = m(x - x_{1})

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).

Match
the equations A, B, C and D with lines L_{1}, L_{2}, L_{3}
and L_{4}, whose graphs are roughly drawn in the given diagram.

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

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

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 = L_{3}

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 = L_{4}

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 = L_{2}

The slope of the line y = 2 is 0.

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

Hence, D = L_{1}

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.

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

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.

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 - y_{1} = m(x - x_{1})

y + 3 = (x - 11)

5y + 15 = -3x + 33

3x + 5y = 18

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.

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

The equation of the given line is

5x - 3y = 4

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)

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.

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 =

(x_{1}, y_{1}) = (0, -5)

Thus, the required equation of the line is

y - y_{1} = m(x - x_{1})

y + 5 = -3(x - 0)

3x + y + 5 = 0

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

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

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.

(i) Slope of PQ =

Equation of the line PQ is given by

y - y_{1} = m(x - x_{1})

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

(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.

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 = , (x_{1}, y_{1}) = (1, 5)

Equation of the line AC is

y - y_{1} = m(x - x_{1})

y - 5 = (x - 1)

2y - 10 = 3x - 3

3x - 2y + 7 = 0

For line BD:

Slope = m = , (x_{1}, y_{1}) = (-1, 2)

Equation of the line BD is

y - y_{1} = m(x - x_{1})

y - 2 = (x + 1)

3y - 6 = -2x - 2

2x + 3y = 4

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.

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

The given line is

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

3y = x - 2

Slope of this line is .

The required line intersects the given line at right angle.

Slope of the required line =

The required line passes through (0, 0) = (x_{1}, y_{1})

The equation of the required line is

y - y_{1} = m(x - x_{1})

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

Thus, the co-ordinates of the point X are .

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.

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

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.

7x + 6y = 71 28x + 24y = 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

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

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) = (x_{1}, y_{1})

The equation of the required line is

y - y_{1} = m(x - x_{1})

y + b = (x - 0)

by + b^{2} = -ax

ax + by + b^{2} = 0

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.

(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 =

(x_{1}, y_{1}) = (0, 0)

The equation of the median OD is

y - y_{1} = m(x - x_{1})

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 - y_{1} = m(x - x_{1})

y + 3 = (x + 5)

5y + 15 = -3x - 15

3x + 5y + 30 = 0

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?

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

Slope of AB = m_{1} =

Equation of line AB is

y - y_{1} = m_{1}(x - x_{1})

y - 3 = (x + 2)

3y - 9 = -x - 2

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

Slope of the given line 3x = y + 1 is 3 = m_{2}.

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.

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).

x cos + y sin = 2

Slope of this line =

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

Let (4, 3) = (x_{1}, y_{1})

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

y - y_{1} = m_{1}(x - x_{1})

y - 3 = (x - 4)

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.

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

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

y =

Slope of this line = m_{1} =

(i) 2x - y + 7 = 0

y = 2x + 7 = 0

Slope of this line = m_{2} = 2

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

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

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.

Slope of BC =

Equation of the line BC is given by

y - y_{1} = m_{1}(x - x_{1})

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 - y_{1} = m_{1}(x - x_{1})

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).

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.

(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 =

(x_{1}, y_{1}) = (2, 3)

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

y - y_{1} = m_{1}(x - x_{1})

y - 3 = (x - 2)

2y - 6 = -x + 2

x + 2y = 8

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.

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

The co-ordinates of point X are

Slope of RX =

(x_{1}, y_{1}) = (-2, -1)

The required equation of the median RX is given by:

y - y_{1} = m_{1}(x - x_{1})

y + 1 = (x + 2)

7y + 7 = 2x + 4

7y = 2x - 3

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.

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.

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.

(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 =

(x_{1}, y_{1}) = (-4, -2)

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

y - y_{1} = m(x - x_{1})

y + 2 = 1(x + 4)

y + 2 = x + 4

y = x + 2

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.

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

(x_{1}, y_{1}) = (-2, 0)

Required equation of the line AB is given by

y - y_{1} = m(x - x_{1})

y - 0 = (x + 2)

2y = 3x + 6

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

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

Also, x-intercept = 4

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

Slope of AB =

(x_{1}, y_{1}) = (4, 0)

Required equation of the line AB is given by

y - y_{1} = m(x - x1)

y - 0 = (x - 4)

2y = -3x + 12

3x + 2y = 12

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.

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 = 60^{o} (Since, tan 60o =)

Using angle sum property in PQR,

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.

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 - y_{1} = m(x - x_{1})

y - 2 = (x - 0)

2y - 4 = 3x

2y = 3x + 4

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.

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

Slope of the line AB =

(x_{1}, y_{1}) = (6, 4)

The equation of the line AB is given by

y - y_{1} = m(x - x_{1})

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).

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.

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

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.

(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

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.

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

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

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)

Given equation of the line L_{1} is y = 4.

(i)Write the slope of the line L_{2} if L_{2} is the bisector of angle O

(ii)Write the coordinates of point P

(iii)Find the equation of L_{2}

(i)

(ii)

(iii)

(i) equation of AB

(ii) equation of CD

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

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 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

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

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.

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.

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.

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.

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 ratio in which Q divides AC.

(iv) Find the equation of the line AC.

i. k.

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

** **

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.

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
- y_{1} = m(x - x_{1})

⇒ y - (-1) = -2(x - 4)

⇒ y + 1 = -2x + 8

⇒ 2x + y = 7

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

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)

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.

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.

(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.

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

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:

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

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

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

The equation of the given line is

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

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

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).

(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.

(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:

(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.

(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

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?

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.

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?

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.

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.

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

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

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 Exercise Ex. 14(B)

Find the slope of the line whose inclination is:

(i) 0 (ii) 30

(iii) 72 30' (iv) 46

(i) Slope = tan 0^{o}
= 0

(ii) Slope = tan 30^{o}
=

(iii) Slope = tan
72^{o} 30' = 3.1716

(iv) Slope = tan 46^{o}
= 1.0355

Find the inclination of the line whose slope is:

(i) 0 (ii)

(iii) 0.7646 (iv) 1.0875

(i) Slope = tan = 0

= 0^{o}

(ii) Slope = tan =

= 60^{o}

(iii) Slope = tan = 0.7646

= 37^{o}
24'

(iv) Slope = tan = 1.0875

= 47^{o}
24'

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)

We know:

Slope =

(i) Slope =

(ii) Slope =

(iii) Slope =

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)

(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

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)

(i) Slope of AB =

Slope of the line perpendicular to AB =

(ii) Slope of AB =

Slope of the line perpendicular to AB = 1

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

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.

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

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.

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.

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

It can be seen that:

Hence, AB AC.

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

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.

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

Since, slope of AB = slope of CD

Therefore AB || CD

Since, slope of BC = slope of DA

Therefore, BC || DA

Hence, ABCD is a parallelogram

(-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.

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

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

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

Hence, PQRS is a parallelogram.

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

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

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

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

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

Slope of AB =

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

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 60^{o} =

Slope of BC = -tan 60^{o} = -

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.

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

Therefore, slope of AB = 0

As CD || AB, 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 45^{o} with the positive direction of x axis.

(ii) The diagonal BD makes an angle of -45^{o} with the positive direction of x axis.

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.

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

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.

(i) Since, BC is perpendicular to AB,

Slope of AB =

(ii) Since, AD is parallel to BC,

Slope of AD = Slope of BC =

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, )

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

Slope of AB =

Inclination of line AB = = 45^{o}

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

Slope of AB =

Inclination of line AB = = 30^{o}

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

Slope of AB =

Inclination of line AB = = 60^{o}

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

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

Slope of AB = Slope of BC

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

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

Slope of AB = Slope of BC

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.

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.

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, k^{2})

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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