Find the ratio in which the line `3x+4y+2=0` divides the distance between the lines `3x+4y+5=0a n d3x=4y-5=0.`

Options

A. `1:2`
B. `3:7`
C. `2:3`
D. `2:5`

Answer

B

Explanation

Let point A`(x_1,y_1)` lies on the line `3x+4y+5=0`, then `3x_1+4y_1+5=0`

Now, perpendicular distance from A to the line
`3x+4y+2=0`
`rArr(|3x_1+4y_1+2|)/(sqrt(9+16))= (|-5-2|)/(sqrt(9+16))=(-7)/(5)`
Let point `B(x_2,y_2)` lies on the ine `3x+4y-5=0` i.e.,`3x_2+4y_2-5=0`
Perpendicular distance form B to the line `3x_4y+2=0`,
`(|3x_2+4y_2+2|)/(sqrt(9+16))= (|+5-2|)/(sqrt(9+16))=(3)/(5)`
Hence, the required ratio is `(3)/(5):(7)/(5)`i.e.,`3:7`.