I found the coordinates of B to be (1,8).

The equation for AB is y = 3x +5

As for the part (c), my way is way too long.

The key is the given area of right triangle BAD = 40 sq.units

That means 40 = (1/2)(AB)(AD)

So find AB and AD.

AB = sqrt[(1-(-1))^2 +(8-2)^2] = sqrt(40)

AD = sqrt[(x -(-1))^2 +(y -2)^2] ......where (x,y) is corner D.

So,

40 = (1/2)sqrt(40)*sqrt[(x+1)^2 +(y-2)^2]

40 / (1/2)sqrt(40) = sqrt[(x+1)^2 +(y-2)^2]

2sqrt(40) = sqrt[(x+1)^2 +(y-2)^2]

160 = (x+1)^2 +(y-2)^2 --------------(i)

The equation of AD:

slope = same slope as BC = -(1/3)

using the (-1,2),

(y -2) = (-1/3)(x -(-1))

(y-2) = (-1/3)(x+1) -----------------(ii)

Substitute that into (i),

160 = (x+1)^2 +[(-1/3)(x+1)]^2

160 = (x+1)^2 +(1/9)(x+1)^2

160 = (x+1)^2 *[1 +1/9]

160 = (x+1)^2 *(10/9)

160 / (10/9) = (x+1)^2

9*16 = (x+1)^2

3*4 = x+1

x = 12 -1 = 11 -----***

And,

(y-2) = (-1/3)(11 +1) = -4

y = -4 +2 = -2

Therefore, (x,y), the corner D, is (11,-2) ---------answer.