Question

a.) Find the area of the triangle with the given vertices A(1,3), B(-3,5), and C with C = 2A.

(b) Use (a) to find the coordinates of the point D such that the quadrilateral ABCD is a parallelogram?

Answer 1

a) Let us find the area $A$ of the triangle with the given vertices $A(1,3), B(-3,5),$ and $C$ with $C = 2A=(2,6).$

Taking into account that the equation of the line $AC$ is $y=3x$ or $y-3x=0,$ we conclude that the distance (altitude) $h$ from the point $B$ to the line $AC$ is equal to $h=\frac{|5-3(-3)|}{\sqrt{1^2+3^2}}=\frac{14}{\sqrt{10}}.$ Since $AC=\sqrt{1^2+3^2}=\sqrt{10},$ we conclude that

$A=\frac{1}2AC\cdot h=\frac{1}2\sqrt{10}\frac{14}{\sqrt{10}}=7$ (sq. units).

b) Let us find the coordinates of the point $D(a,b)$ such that the quadrilateral $ABCD$ is a parallelogram. Let $O(x,y)$ is the point of the diagonals intersection. Then $x=\frac{1+2}2=\frac{3}2,\ y=\frac{3+6}2=\frac{9}2.$ It follows that $\frac{a-3}2=\frac{3}2,\ \frac{b+5}2=\frac{9}2.$ Therefore, $a=6,\ b=4.$

We conclude that $D(6,4).$