Thread: Lenght and gradient.

Originally Posted by helloying

Given the points P(k,3) Q(1,1) and R(6,2), find the possibe values of k if PQ=2QR.

I did it by finding length of 2QR= 2√(1-6)²+(1-2)² and i got 2√26
so PQ=2√26
√(k-1)² +(3-1)² = 2√26
√(k² -2k+1+4)=2√26

how to find the k? the ans is 11 and -9 and i dont know how to get it

Stop here: √(k-1)² +(3-1)² = 2√26
Square both sides and then move 4 to the other side. Can you solve for k now?