intersection of tangents to a curve, equidistance

[red and white kop!]

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?

 

[galactus]

Your tangent line at Q is incorrect. It should be \(\displaystyle y=5x-20\)

The distance should be \(\displaystyle \sqrt{130}\)

 

[Deleted member 4993]

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 <<< dy/dx = x - 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?

.

 

[red and white kop!]

bwaha ok thats stupid, thanks