Question

Perpendicular distance of the line AB $-2x+3y+4=0$ from the point (2, 2) is

• A. $\sqrt{36/13}$
• B. $\sqrt{10}$
• C. $\sqrt{20}$
• D. $\sqrt{50}$

Answer 1

The correct option is A $\sqrt{36/13}$

Formula,

$d=\frac{|ax+by+c|}{\sqrt{a^2+b^2}}$

$-2x+3y+4=0$

$(2,2)$

$d=\frac{|-2\left(2\right)+\left(3\right)\left(2\right)+4|}{\sqrt{(-2{)}^2+3^2}}$

$=\frac{4+6-4}{\sqrt{13}}$

$\therefore d=\frac{6}{\sqrt{13}}$