Question

The equation of the line joining the point (3, 5) to the point of intersection of the lines 4x + y – 1 = 0 and 7x – 3y – 35 = 0 is equidistant from the points (0, 0) and (8, 34).

विकल्प

• True

• False

Answer 1

This statement is True.

Explanation:

Given equations are

4x + y – 1 = 0  ......(i)

And 7x – 3y – 35 = 0   ......(ii)

From equation (i) y = 1 – 4x   ......(iii)

Putting the value of y in equation (ii) we get

7x – 3(1 – 4x) – 35 = 0

⇒ 7x – 3 + 12x – 35 = 0

⇒ 19x – 38 = 0

⇒ x = 2

From equation (iii) we get,

y = 1 – 4 × 2

⇒ y = – 7

The point of intersection is (2, – 7).

Equation of line joining the point (3, 5) to the point (2, – 7) is

y – 5 = `(-7 - 5)/(2 - 3) (x - 3)`

⇒ y – 5 = 12(x – 3)

⇒ y – 5 = 12x – 36

⇒ 12x – y – 31 = 0    .......(iv)

Distance of equation (iv) from the point (0, 0)

= `|(-31)/sqrt((12)^2 + (-1)^2)|`

= `31/sqrt(145)`

Distance of equation (iv) from the point (8, 34) is

= `|(12 xx 8 - 34 - 31)/sqrt((12)^2 + (-1)^2)|`

= `|(96 - 65)/sqrt(145)|`

= `31/sqrt(145)`