**Chapter 5 Straight Line Ex 5.4**

## Chapter 5 Straight Line Ex 5.4

**Question 1.Find the slope, x-intercept, y-intercept of each of the following lines, i. 2x + 3y-6 = 0 ii. 3x-y-9 = 0 iii. x + 2y = 0Solution:**

i. Given equation of the line is 2x + 3y – 6 = 0.

Comparing this equation with ax + by + c = 0,

we get

a = 2, b = 3, c = -6

ii. Given equation of the line is 3x – y – 9 = 0.

Comparing this equation with ax + by + c = 0,

we get

a = 3, b = – 1, c = – 9

iii. Given equation of the line is x + 2y = 0.

Comparing this equation with ax + by + c = 0,

we get

a = 1, b = 2, c = 0

**Question 2.Write each of the following equations in ax + by + c = 0 form.**
i. y = 2x – 4

ii. y = 4

i. y = 2x – 4

∴ 2x – y – 4 = 0 is the equation in ax + by + c = 0 form.

ii. y = 4

∴ 0x + 1y – 4 = 0 is the equation in ax + by + c = 0 form.

∴ 2x – 3y = 0

∴ 2x – 3y + 0 = 0 is the equation in ax + by + c = 0 form.

[Note: Answer given in the textbook is ‘2x – 3y – 6 = 0’. However, as per our calculation it is ‘2x-3y + 0 = 0’.]

**Question 3.Show that the lines x – 2y – 7 = 0 and 2x – 4y + 15 = 0 are parallel to each other.Solution:**

the given lines are parallel to each other.

**Question 4.Show that the lines x – 2y – 7 = 0 and 2x + y + 1 = 0 are perpendicular to each other. Find their point of intersection.Solution:**

the given lines are perpendicular to each other. Consider,

x – 2y – 7 = 0 …(i)

2x + y + 1 =0 …(ii)

Multiplying equation (ii) by 2, we get

4x + 2y + 2 = 0 …(iii)

Adding equations (i) and (iii), we get

5x – 5 = 0

∴ x = 1

Substituting x = 1 in equation (ii), we get

2 + y + 1 = 0

∴ y = – 3

∴ The point of intersection of the given lines is (1,-3).

**Question 5.If the line 3x + 4y = p makes a triangle of area 24 square units with the co-ordinate axes, then find the value of p.Solution:**

Let the line 3x + 4y = p cuts the X and Y axes at points A and B respectively.

3x + 4y = p

**Question 6.Find the co-ordinates of the foot of the perpendicular drawn from the point A(- 2,3) to the line 3x-y -1 = 0.Solution:**

Let M be the foot of perpendicular drawn from

point A(- 2,3) to the line

3x-y- 1 = 0 …(i)

∴ 3(y – 3) = – 1(x + 2)

∴ 3y – 9 = -x – 2

∴ x + 3y – 7 = 0 …………(ii)

The foot of perpendicular i.e., point M, is the point of intersection of equations (i) and (ii).

By (i) x 3 + (ii), we get 10x -10 = 0

∴ x = 1

Substituting x = 1 in (ii), we get

1 + 3y – 7 = 0

∴ 3y = 6

∴ y = 2

∴ The co-ordinates of the foot of the perpendicular Mare (1,2).

**Question 7.Find the co-ordinates of the circumcentre of the triangle whose vertices are A(- 2, 3), B(6, -1), C(4,3),eSolution:**

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

Let F be the circumcentre of AABC.

Let FD and FE be the perpendicular bisectors of the sides BC and AC respectively.

∴ D and E are the midpoints of side BC and AC respectively.

Since FD passes through (5, 1) and has slope 1/2 equation of FD is

∴ 2 (y – 1) = x – 5

∴ 2y – 2 = x – 5

∴ x – 2y – 3 = 0 …(i)

Since both the points A and C have same y co-ordinates i.e. 3,

the given points lie on the line y = 3.

Since the equation FE passes through E(1, 3),

the equation of FE is x = 1. .. .(ii)

To find co-ordinates of circumcentre, we have to solve equations (i) and (ii).

Substituting the value of x in (i), we get

1 – 2y -3 = 0

∴ y = -1

∴ Co-ordinates of circumcentre F ≡ (1, – 1).

**Question 8.Find the co-ordinates of the orthocentre of the triangle whose vertices are A(3, – 2), B(7,6), C (-1,2).Solution:**

Let O be the orthocentre of ∆ABC.

Let AD and BE be the altitudes on the sides BC and AC respectively.

**Question 9.Show that the lines 3 – 4y + 5 = 0, lx – 8y + 5 = 0 and 4JC + 5y – 45 = 0 are concurrent. Find their point of concurrence.Solution:**

The number of lines intersecting at a point are called concurrent lines and their point of intersection is called the point of concurrence. Equations of the given lines are

3x – 4y + 5 = 0 …(i)

7x-8y + 5 = 0 …(ii)

4x + 5y – 45 = 0 …(iii)

By (i) x 2 – (ii), we get

– x + 5 = 0

∴ x = 5

Substituting x = 5 in (i), we get

3(5) – 4y + 5 = 0

∴ -4y = – 20

∴ y = 5

∴ The point of intersection of lines (i) and (ii) is given by (5, 5).

Substituting x = 5 and y = 5 in L.H.S. of (iii), we get

L.H.S. = 4(5) + 5(5) – 45

= 20 + 25 – 45

= 0

= R.H.S.

∴ Line (iii) also passes through (5, 5).

Hence, the given three lines are concurrent and the point of concurrence is (5, 5).

**Question 10.Find the equation of the line whose x-intercept is 3 and which ¡s perpendicular to the line 3x – y + 23 = 0.Solution:**

Slope of the line 3x – y + 23 = 0 is 3.

∴ Slope of the required line perpendicular to

Since the x-intercept of the required line is 3, it passes through (3, 0).

∴ The equation of the required line is ‘

∴ 3y = x + 3

∴ x + 3y = 3

**Question 11.Find the distance of the origin from the line 7x + 24y – 50 = 0.Solution:**

Let p be the perpendicular distance of origin

fromtheline7x + 24y – 50 = 0

Here, a = 7, b = 24, c = -50

**Question 12.Find the distance of the point A(- 2, 3) from the line 12x – 5y – 13 = 0.Solution:**

Let p be the perpendicular distance of the point A(- 2, 3) from the line 12x – 5y – 13 = 0

Here, a = 12, b = – 5, c = – 13, = -2, = 3

**Question 13.Find the distance between parallel lines 4x – 3y + 5 = 0 and 4xr – 3y + 7 = 0.Solution:**

Equations of the given parallel lines are 4x – 3y + 5 = 0 and 4x – 3y + 1 = 0

Here, a = 4, b = – 3, = 5 and = 7

∴ Distance between the parallel lines

**Question 14.Find the distance between the parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.Solution:**

Equations of the given parallel lines are 3x + 2y + 6 = 0 and

∴ Distance between the parallel lines

**Question 15.Find the points on the line x + y – 4 = 0 which are at a unit distance from the line 4JC + 3y = 10.Solution:**

Let P(, ) be a point on the line x + y – 4 = o.

∴ + – 4 = 0

∴ = 4 – x

_{1}…(i)

Also, distance of P from the line 4x + 3y- 10 = 0 is 1

[Note: The question has been modified]

**Question 16.Find the equation of the line parallel to the X-axis and passing through the point of intersection of lines x + y – 2 = 0 and 4x + 3y = 10.Solution:**

Let u = x + y – 2 = 0 and v = 4x + 3y – 10 = 0

Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0.

∴ (x + y – 2) + k(4x + 3y – 10) = 0 …(i)

∴ x + y – 2 + 4kx + 3ky – 10k = 0

∴ x + 4kx + y + 3ky – 2 – 10k = 0

∴ (1+ 4k)x + (1 + 3k)y – 2 – 10k = 0

But, this line is parallel to X-axis.

∴ Its slope = 0

Substituting the value of k in (i), we get

(x + y – 2) + (4x + 3y – 10) = 0

∴ 4(x +y – 2) – (4x + 3y -10 ) = 0

∴ 4x + 4y – 8 – 4x – 3y + 10 = 0

∴ y + 2 = 0, which is the equation of the required line.

[Note: Answer given in the textbook is 5y – 8= 0. However, as per our calculation it is y + 2 = 0.]

**Question 17.Find the equation of the line passing through the point of intersection of lines x + y – 2 = 0 and 2xr – 3y + 4 = 0 and making intercept 3 on the X-axis.Solution:**

Let u ≡ x + y – 2 = 0 and v ≡ 2x – 3y + 4 = 0

Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0.

∴ (x +y – 2) + k(2x – 3y + 4) = 0 …(i)

But, x-intercept of line is 3.

∴ It passes through (3, 0).

Substituting x = 3 and y = 0 in (i), we get

(3 + 0 – 2) + k(6 – 0 + 4) = 0

∴ 1 + 10k = 0

∴ 10(x + y – 2) – (2x – 3y + 4) = 0

∴ 10x + 10y -20 — 2x + 3y-4 = 0

∴ 8x + 13y – 24 = 0, which is the equation of the required line.

**Question 18.If A(4, 3), B(0, 0) and C(2, 3) are the vertices of ΔABC, then find the equation of bisector of angle BAC.Solution:**

Let the bisector of ∠ BAC meets BC at point D.

∴ Point D divides seg BC in the ratio l(AB) : l(AC)

∴ 18 (y – 3) = 6 (x – 4)

∴ 3(y – 3) = x – 4

∴ 3y – 9 = x – 4

∴ x – 3y + 5 = 0

**Question 19.D(- 1, 8), E(4, – 2), F(- 5, – 3) are midpoints of sides BC, CA and AB of AABC. Findi. equations of sides of ΔABC.ii. co-ordinates of the circumcentre of ΔABC.Solution:**

Let A(), B() and C() be the vertices of ΔABC.

Given, points D, E and F are midpoints of sides BC, CA and AB respectively of ΔABC.

∴ 8(y + 13) = 22x

∴ 4(y + 13) = 11x

∴ 11x – 4y – 52 = 0

ii. Here, A(0, – 13), B(- 10, 7), C(8, 9) are the vertices of ΔABC.

Let F be the circumcentre of AABC.

Let FD and FE be perpendicular bisectors of the sides BC and AC respectively.

D and E are the midpoints of side BC and AC.

**Question 20.0(0, 0), A(6, 0) and B(0, 8) are vertices of a triangle. Find the co-ordinates of the incentre of ∆OAB.Solution:**

Let bisector of ∠O meet AB at point D and bisector of ∠A meet BO at point E

∴ Point D divides seg AB in the ratio l(OA): l(OB)

and point E divides seg BO in the ratio l(AB): l(AO)

Let I be the incentre of ∠OAB.

∴ Point D divides AB internally in 6 : 8

i.e. 3 :4

∴ y = x …(i)

Now, by distance formula,

∴ Point E divides BO internally in 10 : 6 i.e. 5:3

∴ -2y = x – 6

∴ x + 2y = 6 …(ii)

To find co-ordinates of incentre, we have to solve equations (i) and (ii).

Substituting y = x in (ii), we get

x + 2x = 6

∴ x = 2

Substituting the value of x in (i), we get

y = 2

∴ Co-ordinates of incentre I ≡ (2, 2).

Alternate Method:

Let I be the incentre.

I lies in the 1st quadrant.

OPIR is a square having side length r.

Since OA = 6, OP = r,

PA = 6 – r

Since PA = AQ,

AQ = 6 – r …(i)

Since OB = 8, OR = r,

BR = 8 – r

∴ BR = BQ

∴ BQ = 8 – r …(ii)

AB = BQ + AQ

∴ BQ + AQ= 10

∴ (8 – r) + (6 – r) = 10

∴ 2r = 14- 10 = 4

∴ r = 2

∴ I = (2,2)