Question

Find the equation of the line passing through $(-3,5)$ and perpendicular to the line through the points $(2,5)(-3,6)$.

• (A) $x+y+20=0$
• (B) $x-y+20=0$
• (C) $x+5y+20=0$
• (D) $5x-y+20=0$

Answer 1

The Correct Option is D

Explanation:
The equation of the line passing through $(x_1,y_1)$ with the slope $m$ is:$(y-y_1)=m(x-x_1)$The slope of line passing through the points $(x_1,y_1)$ and $(x_2,y_2)$ is:$=\frac{y_2-y_1}{x_2-x_1}$The slope of line passing through the points $(2,5)(-3,6)$ is $=\frac{6-5}{-3-2}$$\Rightarrow m_1=\frac{1}{-5}=\frac{-1}{5}$The two non-vertical lines are perpendicular to each other only if and only if their slopes are negative reciprocals of each other.The slope of perpendicular to the line through the points $(2,5)(-3,6)$ is ${\mathrm{m}}_2=\frac{-1}{{\mathrm{m}}_1}$$\Rightarrow {\mathrm{m}}_2=5$The equation of the line passing through $(x_1,y_1)=(-3,5)$ and perpendicular to the line through the points $(2,5)(-3,6)$ is:$(y-y_1)=m(x-x_1)$$\Rightarrow (y-5)=5(x+3)$$\Rightarrow 5x-y+20=0$Hence, the correct option is (D).