Question

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

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

Answer 1

Answer: (A)

(A) 3x+y−2=0

Answer 2

Explanation:
Given,Line x−3y+5=0The general equation of a line is y=mx+cWhere 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⇒y=13x+53⇒ Slope (m1)=13 and c1=53Now for the slope of the perpendicular line m2m1×m2=−1⇒13×m2=−1⇒m2=−3Perpendicular line has the slope −3 and passes through (2,−4)∴ Equation of the perpendicular line is(y−y1)=m(x−x1)⇒y−(−4)=−3(x−2)⇒y+3x−2=0Hence, the correct option is (A).