Question

Three vertices of a rectange are aas (3,2),(-4,2) and (4,-5).Plot these points and find the coordinates of the fourt vertices.

Answer 1

We have next picture of three vertices of the rectangle:

If we take $\mathrm{A} = (3,2)$ , $\mathrm{B} = (-4,2)$ and $\mathrm{C} = (4, -5)$ , then:

Then we can see that it will be parallelogram and we can find vertices D:

Line AD will be parallel to line BC and AB will be parallel to the line CD. So, we can find:

$$A B: y = 2, - 4 \leq x \leq 3$$ $$B C: \frac {x + 4}{8} = \frac {y - 2}{- 7} \Rightarrow y = - \frac {7}{8} \cdot x - \frac {3}{2}$$

So we have that line CD has equation $y = c$ , where $c = \text{const}$ and going through the point $C = (4, -5)$ . So, we can say that line CD has equation: $y = -5$ .

The line AD has equation $y = -\frac{7}{8} \cdot x + c$ , where $c = \text{const}$ and going through the point

$\mathrm{A} = (3,2)$ . So: $y(3) = -\frac{7}{8} \cdot 3 + c = 2 \Rightarrow c = 2 + \frac{21}{8} = \frac{37}{8}$ and line AD has equation:

$$y = - \frac {7}{8} \cdot x + \frac {3 7}{8}.$$

And we can find point D:

$$\begin{array}{l} AD \cap CD: \\ y = -\frac{7}{8} \cdot x + \frac{37}{8} = -5 \Rightarrow -\frac{7}{8} \cdot x = -\frac{77}{8} \Rightarrow x = 11, y = -5 \Rightarrow \\ \Rightarrow D = (11, -5) \end{array}$$

And we have such rectangle:

Answer: $D = (11, -5)$.