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

Question

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 (A) (130/17√29) (B) (13/7√29) (C) (130/7) (D) None of these

Tardigrade Admin · Accepted Answer

On solving the equations 2x - 3y + 5 = 0 and 3x + 4y = 0, we get x=(-20/17), y=(15/17) ∴ Point of intersection is ((-20/17), (15/17)) ∴ Distance =(|5((-20/17))-2((15/17))|/√25+4) =(130/17√29)

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

$\frac{130}{17 29}$

$\frac{13}{7 29}$

$\frac{130}{7}$

None of these

On solving the equations $2 x - 3 y + 5 = 0$ and $3 x + 4 y = 0$ ,
we get $x = \frac{- 20}{17}$ , $y = \frac{15}{17}$
$∴$ Point of intersection is $(\frac{- 20}{17}, \frac{15}{17})$
$∴$ Distance $= \frac{∣ 5 ( \frac{- 20}{17} ) - 2 ( \frac{15}{17} ) ∣}{25 + 4}$
$= \frac{130}{17 29}$

Q. 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

1729​130​

729​13​

7130​