# Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 14 - Equation of a Line

## Selina Textbook Solutions Chapter 14 - Equation of a Line

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## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 14 - Equation of a Line Page/Excercise 14(A)

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.

(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.

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:

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 equation of the given line is

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

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 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) 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

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.

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.

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

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.

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 14 - Equation of a Line Page/Excercise 14(E)

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

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

The equation of the given line is

5x - 3y + 4 = 0

Slope of this line =

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

Slope of the required line =

Thus, the equation of the required line is

y - y1 = m(x - x1)

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

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

(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

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

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

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

Co-ordinates of O are

Slope of AC =

For line AC:

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

Equation of the line AC is

y - y1 = m(x - x1)

y - 5 = (x - 1)

2y - 10 = 3x - 3

3x - 2y + 7 = 0

For line BD:

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

Equation of the line BD is

y - y1 = m(x - x1)

y - 2 = (x + 1)

3y - 6 = -2x - 2

2x + 3y = 4

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) = (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

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

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

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

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

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

43x = 215

x = 5

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

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

4x - 2y = 1

2y = 4x - 1

y = 2x -

Slope of this line = 2

Slope of the required line =

The required equation of the line is

y - y1 = m(x - x1)

y - 6 = (x - 5)

2y - 12 = -x + 5

x + 2y = 17

The given line is

Slope of this line =

Slope of the required line =

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

Putting x = 0 in the equation , we get,

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

The equation of the required line is

y - y1 = m(x - x1)

y + b = (x - 0)

by + b^{2} = -ax

ax + by + b^{2} = 0

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

Co-ordinates of point D are

Slope of OD =

(x1, y1) = (0, 0)

The equation of the median OD is

y - y1 = m(x - x1)

y - 0 = -1(x - 0)

x + y = 0

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

Slope of OA =

Slope of the required altitude =

The equation of the required altitude through B is

y - y1 = m(x - x1)

y + 3 = (x + 5)

5y + 15 = -3x - 15

3x + 5y + 30 = 0

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

Slope of AB = m1 =

Equation of line AB is

y - y1 = m1(x - x1)

y - 3 = (x + 2)

3y - 9 = -x - 2

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

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

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

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

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

x = 1 and y = 2

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

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

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

x cos + y sin = 2

Slope of this line =

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

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

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

y - y1 = m1(x - x1)

y - 3 = (x - 4)

(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

Slope of BC =

Equation of the line BC is given by

y - y1 = m1(x - x1)

y + 2 = (x + 1)

4y + 8 = 3x + 3

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

(i) Slope of line perpendicular to BC =

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

y - y1 = m1(x - x1)

y - 5 = (x - 0)

3y - 15 = -4x

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

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

(1) 9x - 12y = 15

(2) 16x + 12y = 60

Adding the above two equations, we get,

25x = 75

x = 3

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

y = 1

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

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

(ii) Slope of BC =

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

(x1, y1) = (2, 3)

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

y - y1 = m1(x - x1)

y - 3 = (x - 2)

2y - 6 = -x + 2

x + 2y = 8

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

The co-ordinates of point X are

Slope of RX =

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

The required equation of the median RX is given by:

y - y1 = m1(x - x1)

y + 1 = (x + 2)

7y + 7 = 2x + 4

7y = 2x - 3

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.

(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

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

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

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,

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

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).

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

(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

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

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)

(i)

(ii)

(iii)

** **

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

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 14 - Equation of a Line Page/Excercise 14(B)

(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

(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'

We know:

Slope =

(i) Slope =

(ii) Slope =

(iii) Slope =

(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

(i) Slope of AB =

Slope of the line perpendicular to AB =

(ii) Slope of AB =

Slope of the line perpendicular to AB = 1

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.

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.

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.

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

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.

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.

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

Slope of AB =

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} = -

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

Therefore, slope of AB = 0

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

As BC AB, slope of BC =

As AD AB, slope of AD =

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

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

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

(i) Since, BC is perpendicular to AB,

Slope of AB =

(ii) Since, AD is parallel to BC,

Slope of AD = Slope of BC =

(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}

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

Slope of AB = Slope of BC

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

Slope of AB = Slope of BC

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.

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 =

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 14 - Equation of a Line Page/Excercise 14(C)

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.

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.

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.

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.

(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

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).

(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).

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

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

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:

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

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

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

(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

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

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

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

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

(i)

(ii)

(iii)

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

(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})

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

## Selina Concise Mathematics - Part II Solution for Class 10 Mathematics Chapter 14 - Equation of a Line Page/Excercise 14(D)

(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 =

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) 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.

(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 =

(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) 2x - by + 3 = 0

by = 2x + 3

y =

Slope of this line =

ax + 3y = 2

3y = -ax + 2

y =

Slope of this line =

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

(ii) mx + 3y + 7 = 0

3y = -mx - 7

y =

Slope of this line =

5x - ny - 3 = 0

ny = 5x - 3

y =

Slope of this line =

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

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.

(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}

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.

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.

(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) 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

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

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

(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

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

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

(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) 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

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

(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).

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).

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}

## Selina Concise Mathematics X Class 10 Chapter Solutions

- Chapter 1 - Value Added Tax
- Chapter 2 - Banking (Recurring Deposit Accounts)
- Chapter 3 - Shares and Dividends
- Chapter 4 - Linear Inequations (in one variable)
- Chapter 5 - Quadratic Equations
- Chapter 6 - Solving (simple) Problmes (Based on Quadratic Equations)
- Chapter 7 - Ratio and Proportion (Including Properties and Uses)
- Chapter 8 - Remainder And Factor Theorems
- Chapter 9 - Matrices
- Chapter 10 - Arithmetic Progression
- Chapter 11 - Geometric Progression
- Chapter 12 - Reflection (In x-axis, y-axis, x=a, y=a and the origin ; Invariant Points)
- Chapter 13 - Section and Mid-Point Formula
- Chapter 14 - Equation of a Line
- Chapter 15 - Similarity (With Applications to Maps and Models)
- Chapter 16 - Loci (Locus and its Constructions)
- Chapter 17 - Circles
- Chapter 18 - Tangents and Intersecting Chords
- Chapter 19 - Constructions (Circles)
- Chapter 20 - Cylinder, Cone and Sphere (Surface Area and Volume)
- Chapter 21 - Trigonometrical Identities (Including Trigonometrical Ratios of Complementary Angles and Use of Four Figure Trigonometrical Tables)
- Chapter 22 - Heights and Distances
- Chapter 23 - Graphical Representation (Histograms and Ogives)
- Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode)
- Chapter 25 - Probability

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