Thread: Coordinate Geometry

Please do part(c).

I have found the coords. of point B which is (5/4,35/4).

The answer to part (c) is (11,-2)

Originally Posted by maybeline9216

Please do part(c).

I have found the coords. of point B which is (5/4,35/4).

The answer to part (c) is (11,-2)

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.

Ops sorry... i careless=(