Question

Find the value of x such that the area of a triangle whose vertices have coordinates (6, 5), (8, 2) and (x, 11) is 15 square units.

Answer 1

Answer:

$x =12\ or\ -8$

Step-by-step explanation:

Given

$(x_1,y_1) = (6,5)$

$(x_2,y_2) = (8,2)$

$(x_3,y_3) = (x,11)$

$Area = 15$

Required

Find x

The area is calculated as thus:

$Area = \frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Substitute values

$15 = \frac{1}{2}|6*(2 - 11) + 8*(11 - 5) + x(5 - 2)|$

$15 = \frac{1}{2}|6*(-9) + 8*(6) + x(3)|$

$15 = \frac{1}{2}|-54 + 48 + 3x|$

$15 = \frac{1}{2}|-6 + 3x|$

Multiply through by 2

$2 * 15 = \frac{1}{2}|-6 + 3x| * 2$

$30 = |-6 + 3x|$

Remove absolute sign

$30 = -6 + 3x\ or\ -30 = -6 + 3x$

Add 6 to both sides

$36 = 3x\ or\ -24 = 3x$

Divide by 3

$12 = x\ or\ -8 = x$

$x =12\ or\ -8$

So, the coordinates are (-8, 11) or (12,11)