Question

Find the equation of the line perpendicular to the line $x-3y+5=0$ and passes through the point $(2,-4)$.

• (A) $3x+y-2=0$
• (B) $3x-y-10=0$
• (C) $x-3y-14=0$
• (D) $x+3y+10=0$

Answer 1

The Correct Option is A

Explanation:
Given,Line $x-3y+5=0$The general equation of a line is $y=mx+c$Where $m$ is the slope and $c$ is any constantThe slope of parallel lines is equal.The slope of the perpendicular line have their product $=-1$$\Rightarrow y=\frac{1}{3}x+\frac{5}{3}$$\Rightarrow \text{ Slope }(m_1)=\frac{1}{3}\text{ and }{c}_1=\frac{5}{3}$Now for the slope of the perpendicular line $m_2$$m_1\times m_2=-1$$\Rightarrow \frac{1}{3}\times m_2=-1$$\Rightarrow m_2=-3$Perpendicular line has the slope $-3$ and passes through $(2,-4)$$\therefore$ Equation of the perpendicular line is$(y-y_1)=m(x-x_1)$$\Rightarrow y-(-4)=-3(x-2)$$\Rightarrow y+3x-2=0$Hence, the correct option is (A).