Question

The distance of the point $(2,5)$ from the line $3x+y+4=0$ measured parallel to the line $3x-4y+8=0$ is-

• A. 15$/2$
• B. 9$/2$
• C. $5$
• D. None

Answer 1

The correct option is A 15$/2$

Let the equation of the line parallel to $3x-4y+8=0$ is $3x-4y+c=0$ . . . eq.1

The point $(2,5)$ lies on eq.1

$\therefore 3\left(2\right)-4\left(5\right)+c=0\Rightarrow c=14.$

$\therefore$ The equation of the line becomes

$3x-4y+14=0$ . . . . eq.2

Given line is $3x+y+4=0$. . . . eq.3

Solving eq.2 & eq.3 we get the point of intersection of the lines and that is $(-2,2)$

$\therefore$ The required distance is

$d=\sqrt{(2+2{)}^2+(5-2{)}^2}=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}=5$

$\therefore$ The distance is 5 unit.