Parallelogram on graph problem

[Ilikebugs]

View attachment 6377

Is there a way to do this other than guess and check?

 

[Prove It]

Is there a way to do this other than guess and check?

The length of segment AD is $\displaystyle \begin{align*} \sqrt{ \left( 40 - 0 \right) ^2 + \left( 0 - 30 \right) ^2 } = \sqrt{1600 + 900} = \sqrt{2500} = 50 \end{align*}$.

The length of segment AB is $\displaystyle \begin{align*} \sqrt{ \left( k - 0 \right) ^2 + \left( 50 - 30 \right) ^2 } = \sqrt{ k^2 + 400 } \end{align*}$.

The length of segment BD is $\displaystyle \begin{align*} \sqrt{ \left( 40 - k \right) ^2 + \left( 0 - 50 \right) ^2 } = \sqrt{ 1600 - 80\,k + k^2 + 2500 } = \sqrt{ k^2 - 80\,k + 4100 } \end{align*}$.

Thus the area of triangle ABD can be found using Heron's Formula and the length of the parallelogram is double that. Then you can set this quantity equal to 1340 and solve for k.

 

[Greg]

$$\left|\vec{AB}\times\vec{AD}\right|=1340\implies k=18,\quad B(x,y)=(18,50)$$

$$\vec{AB}+\vec{AD}=\vec{AC}\implies C(x,y)=(58,20)$$