Question

If the equation $12x^2+7xy-p{y}^2-18x+qy+6=0$ represents a pair of perpendicular straight lines, then

• A. $p=12,q=1$
• B. $p=1,q=12$
• C. $p=–1,q=12$
• D. $p=1,q=–12$

Answer 1

The correct option is A

$p=12,q=1$

Explanation for correct option

Step 1: Condition used

1. We know that the equation $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$ represents a pair of straight lines if $∆=\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|=0$
2. The pair of lines are perpendicular if $a+b=0$

Step 2: Solve for the value of $p$

Given that equation of the pair of straight line is $12x^2+7xy-p{y}^2-18x+qy+6=0$

Now comparing with the general equation of the pair of straight line we have

$a=12,b=-p,h=\frac{7}{2},g=-9,f=\frac{q}{2},c=6$

From second condition we have the pair of lines are perpendicular if $a+b=0$

$$\begin{gathered} \therefore 12-p=0 \\ \Rightarrow p=12 \end{gathered}$$

Step 3: Solve for the value of $q$

From first condition we get the equation represents a pair of straight line is $∆=\left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|=0$

$\Rightarrow$ $abc+2fgh-a{f}^2-b{g}^2-c{h}^2=0$

$$\begin{gathered} \therefore \left|\begin{array}{ccc}a& h& g\\ h& b& f\\ g& f& c\end{array}\right|=0 \\ \Rightarrow \left|\begin{array}{ccc}12& \frac{7}{2}& -9\\ \frac{7}{2}& -p& \frac{q}{2}\\ -9& \frac{q}{2}& 6\end{array}\right|=0 \\ \Rightarrow 12·\left(-p\right)·6+2·\left(\frac{q}{2}\right)·\left(-9\right)·\left(\frac{7}{2}\right)–12·\left(\frac{q}{2}\right)·2+p·{(-9)}^2–6·{\left(\frac{7}{2}\right)}^2=0 \\ \Rightarrow -72·p–63·\frac{q}{2}–3·q^2+81·p–\frac{147}{2}=0 \\ \Rightarrow 9p-63·\frac{q}{2}–3·q^2–\frac{147}{2}=0 \end{gathered}$$

Put $p=12$

$$\begin{gathered} \Rightarrow 108-63·\frac{q}{2}–3·q^2–\frac{147}{2}=0 \\ \Rightarrow 3q^2+\frac{63q}{2}-\frac{69}{2}=0 \\ \Rightarrow 2q^2+21q-23=0 \\ \Rightarrow 2q^2+23q-2q-23=0 \\ \Rightarrow q\left(2q+23\right)-1\left(2q+23\right)=0 \\ \Rightarrow \left(2q+23\right)\left(q-1\right)=0 \\ \therefore q=1,-\frac{23}{2} \end{gathered}$$

Therefore $p=12,q=1$

Hence option (A) is correct i.e. $p=12,q=1$