Question

The equation of the line through the intersection of the lines 2x – 3y = 0 and 4x – 5y = 2 and

Column C1	Column C2
(a) Through the point (2, 1) is	(i) 2x – y = 4
(b) Perpendicular to the line (ii) x + y – 5 = 0 x + 2y + 1 = 0 is	(ii) x + y – 5 = 0
(c) Parallel to the line (iii) x – y –1 = 0 3x – 4y + 5 = 0 is	(iii) x – y –1 = 0
(d) Equally inclined to the axes is	(iv) 3x – 4y – 1 = 0

Answer 1

Column C1	Column C2
(a) Through the point (2, 1) is	(i) x – y –1 = 0
(b) Perpendicular to the line (ii) x + y – 5 = 0 x + 2y + 1 = 0 is	(ii) 2x – y = 4
(c) Parallel to the line (iii) x – y –1 = 0 3x – 4y + 5 = 0 is	(iii) 3x – 4y – 1 = 0
(d) Equally inclined to the axes is	(iv) x + y – 5 = 0

Explanation:

(a) Given equations are 2x – 3y = 0   ......(i)

And 4x – 5y = 2   ......(ii)

Equations of line passing through eq. (i) and (ii) we get

(2x  – 3y) + k(4x  – 5y  –2) = 0  .....(iii)

If equation (iii) passes through (2, 1), we get

(2 × 2 – 3 × 1) + k(4 × 2 – 5 × 1 – 2) = 0

⇒ (4 – 3) + k(8 – 5 – 2) = 0

⇒ 1 + k(8 – 7) = 0

⇒ k = – 1

So, the required equation is

(2x – 3y) – 1(4x – 5y – 2) = 0

⇒ 2x – 3y – 4x + 5y + 2 = 0

⇒ – 2x + 2y + 2 = 0

⇒ x – y – 1 = 0

(b) Equation of any line passing through the point of intersection of the line 2x – 3y = 0 and 4x – 5y = 2 is

(2x – 3y) + k(4x – 5y – 2) = 0   ......(i)

⇒ (2 + 4k)x + (– 3 – 5k)y – 2k = 0

Slope = `(-(2 + 4k))/(-3 - 5k) = (2 + 4k)/(3 + 5k)`

Slope of the given line x + 2y + 1 = 0 is `- 1/2`.

If they are perpendicular to each other then

`- 1/2((2 + 4k)/(3 + 5k)) = -1`

⇒ `(1 + 2k)/(3 + 5k)` = 1

⇒ 1 + 2k = 3 + 5k

⇒ 3k = – 2

⇒ k = `(-2)/3`

Putting the value of k is eq. (i) we get

`(2x - 3y) - 2/3 (4x - 5y - 2)` = 0

⇒ 6x – 9y – 8x + 10y + 4 = 0

⇒ – 2x + y + 4 = 0

⇒ 2x – y = 4

(c) Given equations are

2x – 3y = 0    .......(i)

4x – 5y = 2  ......(ii)

Equation of line passing through equation (i) and (ii) we get

(2x – 3y) + k(4x – 5y – 2) = 0

⇒ (2 + 4k)x + (– 3 – 5k)y – 2k = 0

Slope = `(-(2 + 4k))/(-3 - 5k) = (2 + 4k)/(3 + 5k)`

Slope of the given line 3x – 4y + 5 = 0 is `3/4`.

If the two equations are parallel, then

`(2 + 4k)/(3 + 5k) = 3/4`

⇒ 8 + 16k = 9 + 15k

⇒ 16k – 15k = 9 – 8

⇒ k = 1

So, the equation of the required line is

(2x – 3y) + 1(4x – 5y – 2) = 0

2x – 3y + 4x – 5y – 2 = 0

⇒ 6x – 8y – 2 = 0

⇒ 3x – 4y – 1 = 0

(d) Given equations are

2x – 3y = 0   ......(i)

4x – 5y – 2 = 0  ......(ii)

Equation of line passing through the intersection of equation (i) and (ii) we get

(2x – 3y) + k(4x – 5y – 2) = 0

⇒ (2 + 4k)x + (– 3 – 5k)y – 2k = 0

Slope = `(2 + 4k)/(3 + 5k)`

Since the equation is equally inclined with axes

∴ Slope = tan 135° = tan(180° – 45°)

= – tan 45° = – 1

So `(2 + 4k)/(3 + 5k) = -1`

⇒ 2 + 4k = – 3 – 5k

⇒ 4k + 5k = – 3 – 2

⇒ 9k = – 5

⇒ k = `(-5)/9`

Required equation is `(2x - 3y) - 5/9 (4x - 5y - 2)` = 0

⇒ 18x – 27y – 20x + 25y + 10 = 0

⇒ – 2x – 2y + 10 = 0

⇒ x + y – 5 = 0