ronit bhatiya
Last Activity: 15 Years ago
Let the equation of perpendicular bisector is y = mx + c
c = -4
So, y = mx - 4 .............(i)
Center point between P(1, 4) and Q(k, 3) will be ((1+k)/2, 7/2)
In equation (i) passes thru ((1+k)/2, 7/2)
So, 7/2 = m((1+k)/2) - 4
7 = m + mk - 8
m + mk - 15 = 0 .................(ii)
m = gradient of the perpendiculat line
m = - (gradient of PQ) = -(1-k)/(4-3) = k - 1
Replace value of m in equation (ii)
k - 1 + k(k - 1) - 15 = 0
k2 = 16
k = ±4