The distance of the point of intersection of the lines `2x-3y+5=0` and `3x+4y=0` from the line `5x-2y=0` is

Options

A. `(130)/(17sqrt129)`
B. `(13)/(7sqrt29)`
C. `(130)/(7)`
D. None of these

Answer

A

Explanation

Given equation of lines
`2x-3y+5=0`
and `3x+4y=0`
From Eq.(ii),put the volume of `x=(-4y)/(3)` in Eq.(i), we get
`2((-4y)/(3))-3y+5=0`
`rArr -8y-9y+15=0`
`rArry=(15)/(17)`
From Eq.(ii), `3x+4.(15)/(17)=0`
`rArr x=(-60)/(17.3)=(-20)/(17)`
So, the point of intersection is `((-20)/(17),(15)/(17))`
`therefore` Required distances form the line `5x-2y=0` is,
`d=|-5xx(20)/(17)-2((15)/(17))|/(sqrt(25+4))=|(-100)/(17)-(30)/(17)|/(sqrt29)=(130)/(17sqrt29)`
`[because distance of a point `p(x_1y_1)` form the line `ax+by+c=0` is `d =|(ax_1+by_1+c)|/(sqrt(a^2+b^2`]`