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

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.

(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.

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 bisects 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 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.

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.

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

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

Slope of AB =

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 (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 =

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 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.

Slope of the line = tan 60^o =

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.

(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

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).

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

Slope =

The equation of the line AB is given by:

y − y₁ = m(x − x₁)

y − 4 = −1(x + 3)

y − 4 = −x − 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).

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 - y₁ = m(x - x₁)

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 - y₁ = m(x - x₁)

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 - 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

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 - 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

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

(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

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

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

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

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

For line AB:

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

(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 - y₁ = m(x - x₁)

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 - y₁ = m(x - x₁)

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

(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 =

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.

(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 + 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, 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.

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.

(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.

(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

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

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

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

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

(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 =

(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

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

(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).

For the line AB:

(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:

(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).

Consider 

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, AL₃

Consider 

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, BL₄

Consider 

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, CL₂

Consider 

The slope of the line y = 2 is 0.

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

Hence, DL₁

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

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)

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

(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

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 + 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

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.

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)

(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.

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).

(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

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 45^o = 1)

Equation of line CD is y = x - 1

Slope of CD =

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

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 =

(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).

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

(i) Slope of AB =

(ii) Slope of perpendicular bisector of AB

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

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 BD

Therefore, the equation of BD is given by

(i) 

Since L₂ is the bisector of angle O,

(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)

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