# Thread: intersection of tangents to a curve, equidistance

1. ## intersection of tangents to a curve, equidistance

It is given that y = 0.5(x^2)  3x + 12
The points P and Q on the graph have x-coordinates 0 and 8 respectively. The tangents at P and Q meet at R. Show that the point (11, 9) is equidistant from P, Q and R.

Alright so this is my work so far:
dy/dx = 2x  3
P (0, 12), Q (8, 20)
Equation of tangent at P: (y  12)/(x  0) = -3 => y = 12 - 3x
Equation of tangent at Q: (y  20)/(x  8) = 13 => y = 13x - 84
Point of intersection R is defined by: 12  3x = 13x  84 => 16x = 96 => x = 6
R(6,-6)

Distance between (11, 9) and P = sqrt((11^2) + (-3^2)) = sqrt(130)
Distance between (11, 9) and Q = sqrt((3^2) + (-11^2)) = sqrt(130)
Distance between (11, 9) and R = sqrt(((11-6)^2) + ((9+6))^2) = sqrt(250)

So theres my problem. Where have I messed up?

2. ## Re: intersection of tangents to a curve, equidistance

The problem is your derivative, it has to be:
$\displaystyle \frac{d}{dx}\left(\frac{1}{2}x^2-3x+12\right)=\frac{1}{2}\cdot 2x-3=x-3$