Search any question & find its solution
Question:
Answered & Verified by Expert
If $a x^2-x y-3 y^2-5 x+20 y+c=0$ represents a pair of lines passing through the point $(2,3)$, then $\mathrm{a}-\mathrm{c}=$
Options:
Solution:
1685 Upvotes
Verified Answer
The correct answer is:
$27$
$a x^2-x y-3 y^2-5 x+20 y+c=0$
It passes through $(2,3)$
$$
\therefore 4 a-6-27-10+60+c=0
$$
$\Rightarrow 4 a+c=-17$ ...(1)
$\begin{aligned} & \Rightarrow c=-17-4 a \\ & \text { Also }\left|\begin{array}{rrr}a & -\frac{1}{2} & -\frac{5}{2} \\ -\frac{1}{2} & -3 & 10 \\ -\frac{5}{2} & 10 & c\end{array}\right|=0 \\ & \Rightarrow a(-3 c-100)+\frac{1}{2}\left(\frac{-c}{2}+25\right)-\frac{5}{2}\left(-5-\frac{15}{2}\right)=0 \\ & \Rightarrow-3 a c-100 a-\frac{c}{4}+\frac{25}{2}+\frac{25}{2}+\frac{75}{4}=0 \\ & \Rightarrow 3 a c+100 a+\frac{c}{4}=\frac{175}{4} \\ & \Rightarrow 12 a(-17-4 a)+400 a-17-4 a=175 \\ & \Rightarrow a^2-4 a+4=0 \Rightarrow(a-2)^2=0 \Rightarrow a=2 \\ & \therefore c=-17-4(2)=-25 \Rightarrow a-c=2+25=27 .\end{aligned}$
It passes through $(2,3)$
$$
\therefore 4 a-6-27-10+60+c=0
$$
$\Rightarrow 4 a+c=-17$ ...(1)
$\begin{aligned} & \Rightarrow c=-17-4 a \\ & \text { Also }\left|\begin{array}{rrr}a & -\frac{1}{2} & -\frac{5}{2} \\ -\frac{1}{2} & -3 & 10 \\ -\frac{5}{2} & 10 & c\end{array}\right|=0 \\ & \Rightarrow a(-3 c-100)+\frac{1}{2}\left(\frac{-c}{2}+25\right)-\frac{5}{2}\left(-5-\frac{15}{2}\right)=0 \\ & \Rightarrow-3 a c-100 a-\frac{c}{4}+\frac{25}{2}+\frac{25}{2}+\frac{75}{4}=0 \\ & \Rightarrow 3 a c+100 a+\frac{c}{4}=\frac{175}{4} \\ & \Rightarrow 12 a(-17-4 a)+400 a-17-4 a=175 \\ & \Rightarrow a^2-4 a+4=0 \Rightarrow(a-2)^2=0 \Rightarrow a=2 \\ & \therefore c=-17-4(2)=-25 \Rightarrow a-c=2+25=27 .\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.