Question

The coordinator of the vertices of the triangle are (-8,8),(-8,-4), and

Answer 1

Answer with Step-by-step explanation:

Complete question:

The coordinates  of the vertices of the triangle are (-8,8),(-8,-4), and. Consider QR the base of the triangle. The measure of the base is b = 18 units, and the measure of the height is h = units. The area of triangle PQR is square units.

Let

P=(-8,8)

Q=(-8,-4)

QR=b=18 units

Height of triangle, h=Length of PQ

Distance formula

$\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using the formula

Height of triangle, h=$\sqrt{(-8+8)^2+(-4-8)^2}=12units$

Area of triangle PQR=$\frac{1}{2}\times base\times height$

Area of triangle PQR=$\frac{1}{2}\times 18\times 12$

Area of triangle PQR=108 square units

Length of QR=18units

Let the coordinates of R(x,y)

$\sqrt{(x+8)^2+(y+4)^2}=18$

$(x+8)^2+(y+4)^2=324$

$x^2+64+16x+y^2+8y+16=324$

$x^2+y^2+16x+8y=324-64-16$

$x^2+y^2+16x+8y=244$   ......(1)

Using Pythagoras theorem

$H=\sqrt{P^2+B^2}$

$H=\sqrt{(18)^2+(12)^2}$

$H=6\sqrt{13}$units

$(6\sqrt{13})^2=(x+8)^2+(y-8)^2$

$x^2+64+16x+y^2+64-16y=468$

$x^2+y^2+16x-16y=468-64-64=340$

$x^2+y^2+16x-16y=340$ .....(2)

Subtract equation (2) from (1) we get

$24y=-96$

$y=-96/24=-4$

Using the value of y in equation (1)

$x^2+16x+16-32=244$

$x^2+16x=244-16+32$

$x^2+16x=260$

$x^2+16x-260=0$

$x^2+26x-10x-260=0$

$x(x+26)-10(x+26)=0$

$(x+26)(x-10)=0$

$x=-26, x=10$

Hence, the coordinate of R (10,-4) or (-26,-4).