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

Equation of a Line Exercise TEST YOURSELF

Solution 1(a)

Correct option: (iii) 6

The line passes through the point (3, 3).

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

5(3) – k(3) + 3 = 0

k = 6

Solution 1(b)

Correct option: (i) 30°

Therefore, Slope =

tan θ = tan 30°

Inclination = θ = 30°

Solution 1(c)

Correct option: (ii) A is true, R is true

Since, we have two straight lines.

To check whether these two lines are parallel or not, we need to find their slopes.

And, if the slopes are equal, the lines will be parallel to each other.

Hence, assertion (A) is true and reason (R) is also true.

Solution 1(d)

Correct option: (ii) A is true, R is true

The lines are:

y = x + 4   …. (I)

    … (II)

Slope of line 1 = m1 = 1 = tan θ1

θ1 = 45°

Slope of line 2 = m2 = = tan θ2

θ2 = 60°

Here, θ1 ≠ θ2

So, the lines are not parallel and thus intersecting.

Hence, A is true and R is also true.

Solution 1(e)

Correct option: (ii) –4

4x + 3y = 84   … (I)

3x + ky + 7 = 0   … (II)

As the lines are perpendicular to each other.

Solution 1(f)

Correct option: (i) (0, 5)

As the line 5x + 3y = 15 meets y-axis at point P.

So, x-coordinate of point P is 0.

As x = 0, we have

5(0) + 3y = 15

y = 5

Hence, P = (0, 5).

Solution 1(g)

Correct option: (iv) –1

Slope of diagonal AC

Diagonals of a rhombus are perpendicular to each other.

Slope of AC × Slope of BD = –1

Solution 1(h)

Correct option: (i) 1

Line AB is perpendicular to x-axis.

As OP bisects angle O.

X’OP = 45°

Inclination of line OP is (180° + 45°)

Slope of line OP = tan (180° + 45°) = tan 45° = 1

Solution 1(i)

Correct option: (i) A= (6, 0) and B = (0, –4)

As the line 2x – 3y = 12 meets x-axis at point A, its y-coordinate is 0.

At y = 0, we have

2x – 3(0) = 12

x = 6

A = (6, 0)

As the line 2x – 3y = 12 meets y-axis at point B, its x-coordinate is 0.

At x = 0, we have

2(0) – 3y = 12

y = –4

B = (0, 4)

Solution 2

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

Solution 3

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

 

The equation of the given line is

5x - 3y = 4

rightwards double arrow straight y space equals space 5 over 3 straight x minus 4 over 3

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)

Solution 4

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

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

Solution 6

 

Using space distance space formula comma space we space have colon
AB equals square root of open parentheses 6 minus 3 close parentheses squared plus open parentheses negative 2 minus 2 close parentheses squared end root equals square root of 9 plus 16 end root equals 5
BC equals square root of open parentheses 2 minus 6 close parentheses squared plus open parentheses negative 5 plus 2 close parentheses squared end root equals square root of 16 plus 9 end root equals 5
Thus comma space AB equals BC
Also comma space Slope space of space AB equals fraction numerator negative 2 minus 2 over denominator 6 minus 3 end fraction equals fraction numerator negative 4 over denominator 3 end fraction
Slope space of space BC equals fraction numerator negative 5 plus 2 over denominator 2 minus 6 end fraction equals fraction numerator negative 3 over denominator negative 4 end fraction equals 3 over 4
Slope space of space AB cross times Slope space of space BC equals negative 1
Thus comma space AB perpendicular BC
Hence comma space straight A comma space straight B comma space straight C space can space be space the space vertices space of space straight a space square.
open parentheses straight i close parentheses space space Slope space of space AB equals fraction numerator negative 2 minus 2 over denominator 6 minus 3 end fraction equals equals Slope space of space CD
Equation space of space the space line space CD space is
straight y minus straight y subscript 1 equals straight m open parentheses straight x minus straight x subscript 1 close parentheses
rightwards double arrow straight y plus 5 equals fraction numerator negative 4 over denominator 3 end fraction open parentheses straight x minus 2 close parentheses
rightwards double arrow 3 straight y plus 15 equals negative 4 straight x plus 8
rightwards double arrow 4 straight x plus 3 straight y equals negative 7.... left parenthesis 1 right parenthesis
Slope space of space BC equals fraction numerator negative 5 plus 2 over denominator 2 minus 6 end fraction equals fraction numerator negative 3 over denominator negative 4 end fraction equals 3 over 4 equals Slope space of space AD
Equation space of space the space line space AD space is
straight y minus straight y subscript 1 equals straight m open parentheses straight x minus straight x subscript 1 close parentheses
rightwards double arrow straight y minus 2 equals 3 over 4 open parentheses straight x minus 3 close parentheses
rightwards double arrow 4 straight y minus 8 equals 3 straight x minus 9
rightwards double arrow 3 straight x minus 4 straight y equals 1... left parenthesis 2 right parenthesis
Now comma space straight D space is space the space point space of space intersection space of space CD space and space AD.
Then comma space multiplying space left parenthesis 1 right parenthesis space by space 4 rightwards double arrow 16 straight x plus 12 straight y equals negative 28
Multiplying space open parentheses 2 close parentheses space by space 3 rightwards double arrow 9 straight x minus 12 straight y equals 3
Adding space the space above space two space equations space we space get comma
25 straight x equals negative 25
rightwards double arrow straight x equals negative 1
So comma space 4 straight y equals 3 straight x minus 1 equals negative 3 minus 1 equals negative 4
rightwards double arrow straight y equals negative 1
Thus comma space the space co minus ordinates space of space point space straight D space are space left parenthesis negative 1 comma space minus 1 right parenthesis.

left parenthesis ii right parenthesis
The space equation space of space line space AD space is space found space in space part space left parenthesis straight i right parenthesis
It space is space 3 straight x minus 4 straight y equals 1 space or space 4 straight y equals 3 straight x minus 1.
Slope space of space AC equals fraction numerator negative 5 minus 2 over denominator 2 minus 3 end fraction equals fraction numerator negative 7 over denominator negative 1 end fraction equals 7
Slope space of space BD equals fraction numerator negative 1 over denominator Slope space of space AC end fraction equals negative 1 over 7
The space equation space of space diagonal space BD space is
straight y minus straight y subscript 1 equals straight m open parentheses straight x minus straight x subscript 1 close parentheses
rightwards double arrow straight y plus 2 equals fraction numerator negative 1 over denominator 7 end fraction open parentheses straight x minus 6 close parentheses
rightwards double arrow 7 straight y plus 14 equals negative straight x plus 6
rightwards double arrow straight x plus 7 straight y plus 8 equals 0

 

Solution 7

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.

Solution 8

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

Solution 9

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 equation 5x - 8y = -23,

8y = 5x + 23

8y = 25 + 23

8y = 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

Solution 10

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

Slope of AB = m1 =

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.


Equation of line AB is

y - y1 = m1(x - x1)

y - 3 = (x + 2)

3y - 9 = -x - 2

 

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

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.

Solution 11

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 = m(x - x1)

y - 3 = (x - 4)

Solution 12

(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

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.

Solution 13

Slope of BC =

Equation of the line BC is given by

y - y1 = m(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 = m(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).

Solution 14

(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 = m(x - x1)

y - 3 = (x - 2)

2y - 6 = -x + 2

x + 2y = 8

Solution 15

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 = fraction numerator 2 plus 2 over denominator negative 4 minus 2 end fraction equals negative 2 over 3

Slope of QC =fraction numerator negative 10 plus 8 over denominator 0 minus 4 end fraction equals 1 half

Thus, PB is not parallel to QC.

Hence, PBCQ is a trapezium.

Solution 16

(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

Solution 17

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

Solution 18

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

Solution 19

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,

Solution 20

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

Solution 21

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).

Solution 22

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

(i) Slope of AB =

(ii) Slope of perpendicular bisector of AB fraction numerator negative 1 over denominator slope space of space AB end fraction equals fraction numerator negative 1 over denominator negative 2 end fraction equals 1 half

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

Solution 23

(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

Solution 24

 

 

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)

Slope of BDequals fraction numerator negative 4 minus 8 over denominator 2 minus 5 end fraction equals fraction numerator negative 12 over denominator negative 3 end fraction equals 4

Therefore, the equation of BD is given by

straight y minus straight y subscript 1 equals straight m open parentheses straight x minus straight x subscript 1 close parentheses
rightwards double arrow straight y minus 8 equals 4 open parentheses straight x minus 5 close parentheses
rightwards double arrow straight y minus 8 equals 4 straight x minus 20
rightwards double arrow 4 straight x minus straight y equals 12

Solution 25

 

(i) angle XOY equals 90 degree

Since Lis the bisector of angle O,

angle POX equals 45 degree
rightwards double arrow Slope space pf space straight L subscript 2 equals tan space 45 degree equals 1

 

(ii) Let the co-ordinates of P be (x, y).

Since P lies on L1, its y-coordinate is 4.

Also, L2 passes through orgin,

 

(iii)

Solution 26

 

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

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 

Equation of a Line Exercise Ex. 14(A)

Solution 2

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) (0, 5)

Substituting x = 0 and y = 5 in x – 2y + 5 = 0,

LHS = 0 – 2(5) + 5 = 0 – 10 + 5 = –5 ≠ RHS

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

 

(iii) (–5, 0)

Substituting x = –5 and y = 0 in x – 2y + 5 = 0,

LHS = –5 – 2(0) + 5 = –5 – 0 + 5 = 0 = RHS

Hence, (–5, 0) lies 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.

Solution 3

(i) The given line is

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

Thus, the given statement is false.

 

(ii) 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.

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

Solution 5

The equation of the given line is

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


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 bisects the join of (5, -2) and (-1, 2).

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:

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

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.

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.

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

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 in x + 5y + 2 = 0, we have

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

 

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

Hence, the given lines are concurrent.

Solution 1(c)

Correct option: (i) no

To check whether the line  passes through the point (2, –4), we need to check whether it satisfies the given equation of a line or not.

Thus, the statement is not true.

Solution 1(d)

Correct option: (iii) 2

x + y = 8  ... (I)

x – y = 0 … (II)

Solving the two equations simultaneously, we get

x = 4, y = 4

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

Since, (4, 4) lies on the line mx – 2y = 0.

m(4) 2(4) = 0

m = 2

Solution 1(e)

Correct option: (ii) 4

As the line x + y = 4 bisects the line segment joining the points A(0, k) and B(4, 0).

So, the mid-point of AB satisfies the equation of the line.

Midpoint of AB

Substituting  in the equation, we get

Solution 1(a)

Correct option: (iv) 8

Point (m, –4) lies on the line x + y = 4.

m 4 = 4

m = 8

Solution 1(b)

Correct option: (i) 2

As the line kx – y = 9 passes through the point (6, 3).

So, (6, 3) satisfies the line kx – y = 9.

k(6) 3 = 9

6k = 12

k = 2

Equation of a Line Exercise Ex. 14(B)

Solution 3

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.

Solution 4

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.

Solution 5

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.

Solution 6

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

Solution 7

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.

Solution 8

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


 

Since slope of PQ = slope of QR, the points P, Q, and R are collinear.

Solution 9

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

Slope of AB =

Solution 10

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

Solution 11

(i) Since, BC is perpendicular to AB,

Slope of AB =

 

(ii) Since, AD is parallel to BC,

Slope of AD = Slope of BC =

Solution 2

(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

Solution 12

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

Slope of AB = Slope of BC

Solution 13

 

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.

Solution 14

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 =

Solution 1(a)

Correct option: (iii) 60°

Slope =

Solution 1(b)

Correct option: (iv)

Slope of a line = 5

Slope of a line perpendicular to the given line

Slope of a line perpendicular to the given line

Solution 1(c)

Correct option: (ii)

Origin is (0, 0).

Slope of a line passing through origin and the point (–3, 4) is

Solution 1(d)

Correct option: (iii) 45°

Slope of a line passing through the points (4, 3) and (5, 4) is

Solution 1(e)

Correct option: (ii) 1

Slope of a line passing through the points (8, –5) and (–4, 7) is

Slope of the line perpendicular to the above line = 1

Equation of a Line Exercise Ex. 14(C)

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.

Solution 3

Given, slope =

The line 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.

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.

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

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).

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)

− 4 = 1(x + 3)

y − 4 = − 3

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).

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

Solution 9

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

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

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 - 7 = 1(x - 4)

y - 7 = x - 4

x - y + 3 = 0

 

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

3x - 2y + 12 = 0

Solution 13

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

Slope of AB =

Slope of line 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

Solution 14

(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

Solution 15

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

Solution 16

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

Solution 17

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

Solution 18

Let AB and CD be two equally inclined lines as shown in the figure.


 

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

Solution 19

(i)

 

 

(ii)

 

 

(iii)

Solution 20

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

Solution 21

(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)

Solution 22

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

Solution 1(a)

Correct option: (i) x-axis

Slope of the line passing through the points (–7, 4) and (5, 4) is

Equation of this line is:

y – 4 = 0(x + 7)

y = 4

It is parallel to x-axis.

Solution 1(b)

Correct option: (ii) y – x = 2

Slope of the line = tan 45° = 1

Equation of the line is:

y – 2 = 1(x – 0)

y 2 = x

y – x = 2

Solution 1(c)

Correct option: (iv) (0, 4)

The line intersects y-axis at point A.

So, x-coordinate of A will be 0.

Putting x = 0 in the equation, we have

5(0) – 2y + 8 = 0

y = 4

Hence, A = (0, 4).

Solution 1(d)

Correct option: (i) x – y = 7

Equation of a line with x-intercept 7 and y-intercept –7 is:

Solution 1(e)

Correct option: (iii) slope = 1 and x-intercept

Equation of a line is given as 5x – 5y = 12

This is in the form x = mx + c.

Thus, slope = 1 and x-intercept

Equation of a Line Exercise Ex. 14(D)

Solution 2

(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 =

Solution 3

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

Solution 4

(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 =

Since the slopes of two lines are not equal, the lines are not parallel to each other.

 

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

Solution 5

(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 =

Solution 6

(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 =

Solution 7

(i) 2x - by + 5 = 0

by = 2x + 5

straight y equals 2 over straight b straight x space plus 5 over straight b

Slope of this line =

 

ax + 3y = 2

3y = -ax + 2

y =

Slope of this line =

Since, the lines are parallel, the slopes of the two lines must be 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.

Solution 8

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 their slopes is -1.

Solution 9

(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

Solution 10

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.

Solution 11

 

 

 

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 equation of a line parallel to y-axis is x = a where 'a' is the x-intercept.

Since the line passes through (-5, 7), the equation of a line parallel to given line x = -5.

That is, x + 5 = 0

Solution 13

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

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

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 is, 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

Solution 16

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

Solution 17

(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

Solution 18

(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 square units

Solution 19

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

Solution 20

(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).

Solution 21

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).

Solution 22

Consider straight A identical to straight y equals 2 straight x

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, Arightwards arrowL3

 

 

Consider straight B identical to straight y minus 2 straight x plus 2 equals 0

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, Brightwards arrowL4

 

Consider straight C identical to 3 straight x plus 2 straight y equals 6

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, Crightwards arrowL2

 

Consider straight D identical to straight y equals 2

The slope of the line y = 2 is 0.

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

Hence, Drightwards arrowL1

Solution 1(a)

Correct option: (ii) x + y = 5

Equation of a line with x-intercept 5 and y-intercept also 5 is:

Solution 1(b)

Correct option: (iii) equal in magnitude with opposite signs

Inclination = 1

tan θ = 1

tan θ = tan 45°

θ = 45°

Thus, its intercepts with both the axes are equal in magnitude with opposite signs.

Solution 1(c)

Correct option: (iii) 2x + y = 3

Equation of a line passing through (1, 1) is:

y – 1 = m(x – 1)   … (I)

Also, y = mx + 3

mx + 3 1 = mx m

m = 2

From (I), we have

y – 1 = –2(x – 1)

2x + y = 3

Solution 1(d)

Correct option: (ii) x – y = 3

x – y = 3 … (I)

x + y = 0 … (II)

Solving the two equations, we get

Also, tan θ = tan 45° = 1

Equation of the required line is

Solution 1(e)

Correct option: (iv) x + y = 12

As the line cuts equal intercepts with both the axes, slope of the line = –1.

Equation of the line is

y – 6 = –1(x – 6)

x + y = 12

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