Question

Find the perpendicular distance from $(2,2)$ to $12x-5y+25=0$

• A. $3$
• B. $4$
• C. $5$
• D. $noneofthese$

Answer 1

Answer: (A)

$3$

Point is $(2,2)$

Equation is $12x-5y+25=0$

we know that the perpendicular distance from $(x_1,y_1)$ to $ax+by+c=0$ is $\frac{|a{x}_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$

Perpendicular distance is $\frac{|12\left(2\right)-5\left(2\right)+25|}{\sqrt{{12}^2+5^2}}$

$=\frac{39}{\sqrt{169}}$

$=\frac{39}{13}=3$