Find the equation of the perpendicular bisector of the line joining P(2, -3) to Q(-5, 1)
A.
8y + 14x + 13 = 0
B.
8y - 14x + 13 = 0
C.
8y - 14x - 13 = 0
D.
8y + 14x - 13 = 0
Correct Answer: 8y - 14x - 13 = 0
Explanation
Given P(2, -3) and Q(-5, 1)
Midpoint = (2+(−5)/2,−3+1/2)
= (−3/2,−1)
Slope of the line PQ = 1−(−3)/−5−2
= −4/7
The slope of the perpendicular line to PQ = −1/(−4/7)
= 7/4
The equation of the perpendicular line: y =7/4x + b
Using a point on the line (in this case, the midpoint) to find the value of b (the intercept).
−1=(7/4)(−3/2) + b
−1+21/8 = 13/8 = b
∴
The equation of the perpendicular bisector of the line PQ is y = 7/4x + 13/8
≡ 8y = 14x + 13 ⟹ 8y − 14x − 13 = 0
Given P(2, -3) and Q(-5, 1)
Midpoint = (2+(−5)/2,−3+1/2)
= (−3/2,−1)
Slope of the line PQ = 1−(−3)/−5−2
= −4/7
The slope of the perpendicular line to PQ = −1/(−4/7)
= 7/4
The equation of the perpendicular line: y =7/4x + b
Using a point on the line (in this case, the midpoint) to find the value of b (the intercept).
−1=(7/4)(−3/2) + b
−1+21/8 = 13/8 = b
∴
The equation of the perpendicular bisector of the line PQ is y = 7/4x + 13/8
≡ 8y = 14x + 13 ⟹ 8y − 14x − 13 = 0
