(a) From y = (x${}^2$ - x + 1)(x - 2) - x + 1)(x - 2)
= x${}^3$- 3x${}^2$ + 3x - 2
dy/dx = 3x${}^2$ - 6x + 3
=x${}^2$ - 2x + 1
(x - 1) = 0 twice ; x = 1
y = (1${}^2$ - 1 + 1) (1-2) ; y = -1
Gradient m${}_1$ of tangent = y/x = -1/1 = -1
Using m${}_1$m${}_2$ = -1
Gradient, m${}_2$ of normal =1
Using y-y${}_1$ = m${}_2$(x - x${}_1$)
y + 1 = x - 1; x - y - 2 = 0

x${}_1$,y${}_1$ = ( -1, 2), , x${}_2$,y${}_2$, =(3, -4)
Mid point of PR =$\frac{3-1,-4+2}{2,2}$
= (1-1)
Using $\frac{y_2-y_1}{x_2-x_1}$ = $\frac{y-y_1}{x-x_1}$
$\frac{1--1}{5-1}$ = $\frac{y-1}{x-5}$;
$\frac{2}{4}$ = $\frac{y-1}{x-5}$
2(y-1) = x - 5; 2y - 2 = x - 5
2y - x = -3
x - 2y -3 =0