Question

Points P(1,2) Q(-2,8) and R(h, 14) are collinear. Find the value of h.​

Answer 1

Answer:

Solution

Correct option is

C

−1

The given points are A(−5,1), B(1,p) and C(4,−2)

We have (x1=−5,y1=1),(x2=1,y2=p) and (x3=4,y3=−2)

The given points A,B and C are collinear

Therefore, x1(y2−y3)+x2(y3−y1)+x3(y1−y2)=0

⇒(−5)⋅(p+2)+1⋅(−2−1)+4⋅(1−p)=0

⇒(−5p−10−3+4−4p)=0

⇒−9p=−9

⇒p=−1

Hence, p=−1