The diagonals of a rectangular are along $y+7x-6=0$ and $ y-7x+5=0$. If the area of the rectangle is 8, then find the vertices of the rectangle.
I found the baricentre of the rectangle which is $(\frac{11}{14},\frac{1}{2})$ and then tried finding the diagonal length but it only got more complicated. Please help me.
On
Easy to see that $M\left(\frac{11}{14},\frac{1}{2}\right)$ is a center of the rectangle $ABCD$.
Let $\alpha$ is an angle between diagonals. Thus,
$$\tan\alpha=\frac{14}{49-1}=\frac{7}{24}$$ Hence, since $1+\cot^2\alpha=\frac{1}{\sin^2\alpha}$, we obtain: $$1+\frac{576}{49}=\frac{1}{\sin^2\alpha},$$ which gives $\sin\alpha=\frac{7}{25}$.
Now, $S_{\Delta AMB}=\frac{1}{4}S_{ABCD}=2$.
Thus, $\frac{1}{2}AM^2\sin\alpha=2$, which gives $AM=\frac{10}{\sqrt7}$.
Now, let $y=7x-5$ is an equation of $AC$ and $y=-7x+6$ is an equation of $BD$.
Thus, for the vertexes $A$ and $C$ we get the following equation: $$\left(x-\frac{11}{14}\right)^2+\left(7x-5-\frac{1}{2}\right)^2=\frac{100}{7}$$ or $$(14x-11)^2=56,$$ which gives very ugly numbers.
The equation for the vertexes $B$ and $D$ is $$\left(x-\frac{11}{14}\right)^2+\left(-7x+6-\frac{1}{2}\right)^2=\frac{100}{7},$$ which is similar.
On
** just a hint**
Let $a=11/14,$
$b=1/2$,
length $=2L,$
width $=2l .$
and area=$4Ll=8.$
the vetrices will have coordinates $$(a-L,b-l),(a-L,b+l),(a+L,b-l),(a+L,b+l) .$$
$(a-L,b-l) $ is in the line $:y-7x+5=0$ thus
$$b-l-7 (a-L)+5=0$$ or
$$1/2-2/L-11/2+7L+5=0$$ $$\implies L^2=2/7$$
$$\implies l^2=4/L^2=14$$
The slopes of the two diagonals are $7$ and $-7$ respectively. So the horizontal side and the vertical side of the rectangle are in the ratio of $1:7$. If the horizontal side is $2k$, then the vertical side is $14k$ and
$$(2k)(14k)=8$$
So $k=\frac{\sqrt{14}}{7}$.
The four vertices are $\displaystyle \left(\frac{11}{14}+\frac{\sqrt{14}}{7},\frac{1}{2}+\sqrt{14}\right)$, $\displaystyle \left(\frac{11}{14}+\frac{\sqrt{14}}{7},\frac{1}{2}-\sqrt{14}\right)$, $\displaystyle \left(\frac{11}{14}-\frac{\sqrt{14}}{7},\frac{1}{2}+\sqrt{14}\right)$ and $\displaystyle \left(\frac{11}{14}-\frac{\sqrt{14}}{7},\frac{1}{2}-\sqrt{14}\right)$.