**Chapter 5 Straight Line Ex 5.3**

## Chapter 5 Straight Line Ex 5.3

**Question 1.Write the equation of the line:i. parallel to the X-axis and at a distance of 5 units from it and above it.ii. parallel to the Y-axis and at a distance of 5 units from it and to the left of it.iii. parallel to the X-axis and at a distance of 4 units from the point (- 2,3).Solution:**

i. Equation of a line parallel to X-axis is y = k. Since the line is at a distance of 5 units above X-axis, k = 5

∴ The equation of the required line is y = 5.

ii. Equation of a line parallel to Y-axis is x = h. Since the line is at a distance of 5 units to the left of Y-axis, h = -5

∴ The equation of the required line is x = -5.

[Note: Answer given in the textbook is ‘y = -5

However, we found that ‘x = – 5’.]

iii. Equation of a line parallel to the X-axis is of the form y = k (k > 0 or k < 0).

Since the line is at a distance of 4 units from the point (- 2, 3),

k = 4 + 3 = 7 or k = 3- 4 = -1

∴ The equation of the required line is y = 1 or y = – 1.

**Question 2.Obtain the equation of the line:i. parallel to the X-axis and making an intercept of 3 units on the Y-axis.ii. parallel to the Y-axis and making an intercept of 4 units on the X-axis.Solution:**

i. Equation of a line parallel to X-axis with y-intercept ‘k’ isy = k.

Here, y-intercept = 3

∴ The equation of the required line is y = 3.

ii. Equation of a line parallel to Y-axis with x-intercept ‘h’ is x = h.

Here, x-intercept = 4

∴ The equation of the required line is x = 4.

**Question 3.Obtain the equation of the line containing the point:i. A(2, – 3) and parallel to the Y-axis.ii. B(4, – 3) and parallel to the X-axis.Solution:**

i. Equation of a line parallel to Y-axis is of the form x = h.

Since the line passes through A(2, – 3), h = 2

∴ The equation of the required line is x = 2.

ii. Equation of a line parallel to X-axis is of the formy = k.

Since the line passes through B(4, – 3), k = -3

∴ The equation of the required line is y = – 3.

**Question 4.Find the equation of the line:i. passing through the points A(2, 0) and B(3,4)ii. passing through the points P(2, 1) and Q(2,-1)Solution:**

ii. The required line passes through the points P(2, 1) and Q(2,-1).

Since both the given points have same

x co-ordinates i.e. 2,

the given points lie on the line x = 2.

∴ The equation of the required line is x = 2.

**Question 5.Find the equation of the line:i. containing the origin and having inclination 60°.ii. passing through the origin and parallel to AB, where A is (2,4) and B is (1,7).iii. having slope 1/2 and containing the point (3, -2)iv. containing the point A(3, 5) and having slope 2/3v. containing the point A(4, 3) and having inclination 120°.vi. passing through the origin and which bisects the portion of the line 3JC + y = 6 intercepted between the co-ordinate axes.Solution:**

i. Given, Inclination of line = θ = 60°

Slope of the line (m) = tan θ = tan 60°

= √3

Equation of the line having slope m and passing through origin (0, 0) is y = mx.

.‘. The equation of the required line is y = √3 x

ii. Given, A (2, 4) and B (1, 7)

Since the required line is parallel to line AB, slope of required line (m) = slope of AB

∴ m = – 3 and the required line passes through the origin.

Equation of the line having slope m and passing through origin (0, 0) is y = mx.

∴ The equation of the required line is y = – 3x

iii. Given, slope(m) = 1/2 and the line passes through (3, – 2).

Equation of the line in slope point form is

y-= m(x-)

∴ The equation of the required line is

[y-(- 2)]=1/2 (x-3)

∴ 2(y + 2)=x – 3

∴ 2y + 4 = x – 3

∴ x – 2y – 7 = 0

iv. Given, slope(m) = 2/3 and the line passes through (3, 5).

Equation of the line in slope point form is y- = m(x -)

∴ The equation of the required line is y – 5 = 2/3 (x-3)

∴ 3 (y – 5) = 2 (x – 3)

∴ 3y – 15 = 2x – 6

∴ 2x – 3y + 9 = 0

v. Given, Inclination of line = θ = 120°

Slope of the line (m) = tan θ = tan 120°

= tan (90° + 30°)

= – cot 30°

= – √3

and the line passes through A(4, 3).

Equation of the line in slope point form is y- = m(x -)

∴ The equation of the required line is

y- 3 = –√3(x-4)

∴ y – 3 = –√3 x + 4√3

∴ √3x + y – 3 -4√3 = 0

vi.

Alternate Method:

Given equation of the line is 3x + y = 6 …(i)

Substitute y = 0 in (i) to get a point on X-axis.

∴ 3x + 0 = 6

∴ x = 2

Substitute x = 0 in (i) to get a point on Y-axis.

∴ 3(0) + 7 = 6

∴ y = 6

∴ The line 3x + y = 6 intersects the X-axis and Y-axis at A(2,0) and B(0,6) respectively.

Let M be the midpoint of AB.

Equation of OM is of the formy = mx.

∴ The equation of the required line is y = 3x

∴ 3x – y = 0

**Question 6.Line y = mx + c passes through the points A(2,1) and B(3,2). Determine m and c.Solution:**

Alternate Method:

Points A(2, 1) and B(3, 2) lie on the line y = mx + c.

∴ They must satisfy the equation.

∴ 2m + c = 1 …(i)

and 3m + c = 2 …(ii)

equation (ii) – equation (i) gives m = 1

Substituting m = 1 in (i), we get 2(1) + c = 1

∴ c = 1 – 2 = – 1

**Question 7.Find the equation of the line having inclination 135° and making x-intercept 7.Solution:**

Given, Inclination of line = 0 = 135°

∴ Slope of the line (m) = tan 0 = tan 135°

= tan (90° + 45°)

= – cot 45° = – 1 x-intercept of the required line is 7.

∴ The line passes through (7, 0).

Equation of the line in slope point form is y – = m(x – )

∴ The equation of the required line is y — 0 = – 1 (x – 7)

∴ y = -x + 7

∴ x + y – 7 = 0

**Question 8.The vertices of a triangle are A(3, 4), B(2, 0) and C(- 1, 6). Find the equations of the lines containingi. side BCii. the median ADiii. the midpoints of sides AB and BC.Solution:**

Vertices of AABC are A(3, 4), B(2, 0) and C(- 1, 6).

i. Equation of the line in two point form is

∴ – 3y = 6x – 12

∴ 6x + 3y – 12 = 0

∴ 2x + y – 4 = 0

ii. Let D be the midpoint of side BC.

iii. Let D and E be the midpoints of side AB and side BC respectively.

The equation of the line DE is

∴ -4(y-2) = 2x-5

∴ 2x + 4y – 13 = 0

**Question 9.**

where x-intercept = a, y-intercept = b

∴ x-intercept = – 6 and y-intercept = 4

**Question 10.Find equations of the line which contains the point A(l, 3) and the sum of whose intercepts on the co-ordinate axes is zero.Solution:**

**Question 11.Find equations of the line containing the point A(3, 4) and making equal intercepts on the co-ordinate axes.Solution:**

**Question 12.Find the equations of the altitudes of the triangle whose vertices are A(2, 5), B(6, – 1 ) and C(- 4, – 3).Solution:**

A(2, 5), B(6, – 1), C(- 4, – 3) are the vertices of ∆ABC.

Let AD, BE and CF be the altitudes through the vertices A, B and C respectively of ∆ABC.

∴ Slope of AD = -5 …[∵AD ⊥ BC]

Since altitude AD passes through (2, 5) and has slope – 5,

equation of the altitude AD is y – 5 = -5 (x – 2)

∴ y – 5 = – 5x + 10

∴ 5x +y -15 = 0

**Question 13.Find the equations of perpendicular bisectors of sides of the triangle whose vertices are P(-1, 8), Q(4, – 2) and R(- 5, – 3).Solution:**

Let A, B and C be the midpoints of sides PQ, QR and PR respectively of APQR.

A is the midpoint of side PQ.

∴ 2y + 5 = -18x – 9

∴ 18x + 2y + 14 = 0

∴ 9x + y + 7 = 0

C is the midpoint of side PR.

∴ 11(2y – 5) = – 8 (x + 3)

∴ 22y – 55 = – 8x – 24

∴ 8x + 22y -31 = 0

**Question 14.Find the co-ordinates of the orthocentre of the triangle whose vertices are A(2, – 2), B(l, 1) and C(-1,0).Solution:**

Let O be the orthocentre of AABC.

Let AM and BN be the altitudes of sides BC and AC respectively.

Slope of AM = -2 ,..[∵ AM ⊥ BC]

Since AM passes through (2, – 2) and has slope -2,

equation of the altitude AM is y – (- 2) = – 2 (x – 2)

∴ y + 2 = -2x + 4

∴ 2x + y – 2 = 0 …(i)

**Question 15.N(3, – 4) is the foot of the perpendicular drawn from the origin to line L. Find the equation of line L.Solution:**