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
[tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Using the formula
Height of triangle, h=[tex]\sqrt{(-8+8)^2+(-4-8)^2}=12units[/tex]
Area of triangle PQR=[tex]\frac{1}{2}\times base\times height[/tex]
Area of triangle PQR=[tex]\frac{1}{2}\times 18\times 12[/tex]
Area of triangle PQR=108 square units
Length of QR=18units
Let the coordinates of R(x,y)
[tex]\sqrt{(x+8)^2+(y+4)^2}=18[/tex]
[tex](x+8)^2+(y+4)^2=324[/tex]
[tex]x^2+64+16x+y^2+8y+16=324[/tex]
[tex]x^2+y^2+16x+8y=324-64-16[/tex]
[tex]x^2+y^2+16x+8y=244[/tex] ......(1)
Using Pythagoras theorem
[tex]H=\sqrt{P^2+B^2}[/tex]
[tex]H=\sqrt{(18)^2+(12)^2}[/tex]
[tex]H=6\sqrt{13}[/tex]units
[tex](6\sqrt{13})^2=(x+8)^2+(y-8)^2[/tex]
[tex]x^2+64+16x+y^2+64-16y=468[/tex]
[tex]x^2+y^2+16x-16y=468-64-64=340[/tex]
[tex]x^2+y^2+16x-16y=340[/tex] .....(2)
Subtract equation (2) from (1) we get
[tex]24y=-96[/tex]
[tex]y=-96/24=-4[/tex]
Using the value of y in equation (1)
[tex]x^2+16x+16-32=244[/tex]
[tex]x^2+16x=244-16+32[/tex]
[tex]x^2+16x=260[/tex]
[tex]x^2+16x-260=0[/tex]
[tex]x^2+26x-10x-260=0[/tex]
[tex]x(x+26)-10(x+26)=0[/tex]
[tex](x+26)(x-10)=0[/tex]
[tex]x=-26, x=10[/tex]
Hence, the coordinate of R (10,-4) or (-26,-4).
