Question

Column C1	Column C2
(a) The coordinates of the points P and Q on the line x + 5y = 13 which are at a distance of 2 units from the line 12x – 5y + 26 = 0 are	(i) (3, 1), (–7, 11)
(b) The coordinates of the point on the line x + y = 4, which are at a  unit distance from the line 4x + 3y – 10 = 0 are	(ii) `(- 1/3, 11/3), (4/3, 7/3)`
(c) The coordinates of the point on the line joining A (–2, 5) and B (3, 1) such that AP = PQ = QB are	(iii) `(1, 12/5), (-3, 16/5)`

Answer 1

Column C1	Column C2
(a) The coordinates of the points P and Q on the line x + 5y = 13 which are at a distance of 2 units from the line 12x – 5y + 26 = 0 are	(i) `(1, 12/5), (-3, 16/5)`
(b) The coordinates of the point on the line x + y = 4, which are at a  unit distance from the line 4x + 3y – 10 = 0 are	(ii) (3, 1), (–7, 11)
(c) The coordinates of the point on the line joining A (–2, 5) and B (3, 1) such that AP = PQ = QB are	(iii) `(- 1/3, 11/3), (4/3, 7/3)`

Explanation:

(a) Let P(x₁, y₁)be any point on the given line x + 5y = 13

∴ x₁ + 5y₁ = 13

Distance of line 12x – 5y + 26 = 0 from the point P(x₁, y₁)

2 = `|(12x_1 - 5y_1 + 26)/sqrt((12)^2 + (-5)^2)|`

⇒ 2 = `|(12x_1 - (13 - x_1) + 26)/13|`

⇒ 2 = `|(12x_1 - 13 + x_1 + 26)/13|`

⇒ 2 = `|(13x_1 + 13)/13|`

⇒ 2 = ± (x₁ + 1)

⇒ 2 = x₁ + 1

⇒ x₁ = 1   ......(Taking (+) sign)

And 2 = – x₁ – 1

⇒ x₁ = – 3  ......(Taking (–) sign)

Putting the values of x1 in equation x₁ + 5y₁ = 13.

We get y₁ = `12/5` and `16/5`.

So, the required points are `(1, 12/5)` and `(-3, 16/5)`.

(b) Let P(x₁, y₁) be any point on the given line x + y = 4

∴ x₁ + y₁ = 4   ......(i)

Distance of the line 4x + 3y – 10 = 0 from the point P(x₁, y₁)

1 = `|(4x_1 + 3y_1 - 10)/sqrt((4)^2 + (3)^2)|`

⇒ 1 = `|(4x_1 + 3(4 - x_1) - 10)/5|`

⇒ 1 = `|(4x_1 + 12 - 3x_1 - 10)/5|`

⇒ 1 = `|(x_1 + 2)/5|`

⇒ 1 = `+- ((x_1 + 2)/5)`

⇒ `(x_1 + 2)/5` = 1   ......(Taking (+) sign)

⇒ x₁ + 2 = 5

⇒ x₁ = 3

And `(x_1 + 2)/5 = - 1`  ......(Taking (–) sign)

_⇒ x₁ + 2 = 5

⇒ x₁ = 3

Putting the values of x₁ in equation (i) we get

x₁ + y₁ = 4

At x₁ = 3, y₁ = 1

At x₁ = – 7, y₁ = 11

So, the required points are (3, 1) and (– 7, 11).

(c) Given that AP = PQ = QB

Equation of line joining A(– 2, 5) and B(3, 1) is

y – 5 = `(1 - 5)/(3 + 2) (x + 2)`

⇒ y – 5 = `(-4)/5 (x + 2)`

⇒ 5y – 25 = – 4x – 8

⇒ 4x + 5y – 17 = 0

Let P(x₁, y₁) and Q(x₂, y₂) be any two points on the line AB

P(x₁, y₁) divides the line AB in the ratio 1 : 2

∴ x₁ = `(1.3 + 2(-2))/(1 + 2)`

= `(3 - 4)/3`

= `(-1)/3`

y₁ = `(1.1 + 2.5)/(1 + 2)`

= `(1 + 10)3`

= `11/3`

So, the coordinates of P(x₁, y₁) = `((-1)/3, 11/3)`

Now point Q(x₂, y₂) is the mid-point of PB

∴ x₂ = `(3 - 1/3)/2 = 4/3`

y₂ = `(1 + 11/3)/2 = 7/3`

Hence, the coordinates of Q(x₂, y₂) = `(4/3, 7/3)`