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

`√((x_2-x_1)^2+(y_2-y_1)^2)`

Using the formula

Height of triangle, h=
`√((-8+8)^2+(-4-8)^2)=12units`

Area of triangle PQR=
`(1)/(2)* base* height`

Area of triangle PQR=
`(1)/(2)* 18* 12`

Area of triangle PQR=108 square units

Length of QR=18units

Let the coordinates of R(x,y)

`√((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=√(P^2+B^2)`

`H=√((18)^2+(12)^2)`

`H=6√(13)`units

`(6√(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).