Question

If the lines $3x+y=4$, $x-ay+7=0$ and $bx+2y+5=0$ form three consecutive sides of a rectangle, find the value of a and b.

Answer 1

Given lines are :

$3x+y=4$...(i)

$x-ay+7=0$ .... (ii)

$bx+2y+5=0$ .... (iii)

It's said that these lines form three consecutive sides of a rectangle.

So,

Line $\left(i\right)$ and $\left(ii\right)$ must be perpendicular

Also, line $\left(ii\right)$ and $\left(iii\right)$ must be perpendicular

We know that, for two perpendicular lines the product of their slopes will be $-1$.

Now,

Slope of line $\left(i\right)$ is

$3x+y=4\Rightarrow y=-3x=4$

Hence, slope $\left(m_1\right)=-3$

And, slope of line $\left(ii\right)$ is

$x-ay+7=0\Rightarrow ay=x+7$

$y=\left(\frac{1}{a}\right)x+\frac{7}{a}$

Hence, slope $\left(m_2\right)=\frac{1}{a}$

Finally, the slope of line $\left(iii\right)$ is

$bx+2y+5=0\Rightarrow 2y=-bx-5$

$y=\left(\frac{-b}{2}\right)x-\frac{5}{2}$

Hence, slope $\left(m_3\right)=\left(\frac{-b}{2}\right)$

As line (i), (ii) and (iii) are consecutive sides of rectangle, we have

$m_1\times m_2=-1$ and $m_2\times m_3=-1$

$(-3)\times \left(\frac{1}{a}\right)=-1$ and $\left(\frac{1}{a}\right)\times \left(\frac{-b}{2}\right)=-1$

$-3=-a$ and $\frac{-b}{2a}=-1$

$a=3$ and $b=2a\Rightarrow b=2\left(3\right)=6$

Thus, the value of a is 3 and the value of b is $6$.