Question

The perpendicular distance from the point of intersection of the lines $3x+2y+4=0$, $2x+5y-1=0$ to the line $7x+24y-15=0$ is

• A. $\frac{2}{3}$
• B. $\frac{1}{7}$
• C. $\frac{1}{5}$
• D. $\frac{2}{5}$

Answer 1

Answer: (C)

$\frac{1}{5}$

Point of intersection of $3x+2y+4=0$ & $2x+5y-1=0$ is

$6x+4y+8=0$

$-6x-15y+3=0$

$-11y+11=0$

$y=1$ and $x=-2$

The point of intersection is $(-2,1).$

So, distance of $(7x+24y-15=0)$ line from $(-2,1)$ is

$d=\left|\left(\frac{-7\times 2+24-15}{25}\right)\right|$

$=\left|\left(\frac{10-15}{25}\right)\right|$

$=\left|\frac{5}{25}\right|$

$d=\frac{1}{5}.$