Question

The equation to the diagonals of a rectangle are $y+8x-17=0$ and $y-8x+7=0$. If the area of the rectangle is $8sq.units$, find the equation of the sides of the rectangle.

• A. $y=1$
• B. $y=9$
• C. $x=1$
• D. $x=2$

Answer 1

Answer: (A), (B)

The correct options are
A $y=1$
B $y=9$

Eq of diagonal are

$y+8x-17=0$

$y-8x+7=0$

point of intersection $(3/2,5)$

$m_1=-8$ $△PCD=△PCB=△PAB=△PAD$

$m_2=8$

are of $△PCD$

$1/2\times PM\times CD=1/4\times \left(18\right)=289units$

$1/2\times l\cos \theta \times 2l\sin \theta =2$

$2l^2\cos \theta \sin \theta =4$

$l^2\left(\sin 2\theta \right)=4$

$\tan 2\theta =\left(\frac{m_2-m_1}{1+m_1{m}_2}\right)=\left|\frac{8-(-8)}{1+8(-8)}\right|=\frac{16}{63}$

$\sin 2\theta =\frac{16}{65}$

$l^2=\frac{4\times 65}{16}l=\frac{\sqrt{65}}{2}$

$PM=1/2\left(l\cos \theta \right)$

$=1/2l\sqrt{\frac{1+\cos 2\theta }{2}}$

$=\frac{\sqrt{65}}{2}\sqrt{\frac{1+62/65}{2}}$

$=\frac{\sqrt{65}}{2}.\sqrt{\frac{64}{\sqrt{65}}}=4$

$PM=4$

$PN=l\sin \theta =l\sqrt{\frac{1-\cos 2\theta }{2}}=\frac{\sqrt{65}}{2}\sqrt{\frac{1-\frac{63}{65}}{2}}$

$\Rightarrow 1/2$

$PN=MC=1/2$

Angular bisector$=$

$\frac{y+8x-17}{\sqrt{1^2+8^2}}=\left|\frac{y+8x-17}{\sqrt{1^2+8^2}}\right|$

$x=3/2$ and $y=5$

other can be $x=3/2+1/2$

$x=2$ or $x=1$

$y=5$

$y=5\pm 4$

$y=1$ or $y=9$