MHT CET · Maths · Straight Lines
If \(\mathrm{O}(0,0), \mathrm{A}(1,2)\) and \(\mathrm{B}(3,4)\) are the vertices of triangle OAB , then the joint equation of the altitude and median drawn from O is
- A \(3 x^2-x y-2 y^2=0\)
- B \(3 x^2+x y+2 y^2=0\)
- C \(3 x^2-x y+2 y^2=0\)
- D \(3 x^2+x y-2 y^2=0\)
Answer & Solution
Correct Answer
(D) \(3 x^2+x y-2 y^2=0\)
Step-by-step Solution
Detailed explanation
OD is the median
\(\begin{aligned}
\therefore \quad & D \equiv\left(\frac{1+3}{2}, \frac{2+4}{2}\right) \\
& \Rightarrow D \equiv(2,3)
\end{aligned}\)

Equation of OD is \(y=m x\)
\(\begin{aligned}
& \Rightarrow y=\frac{3}{2} x \\
& \Rightarrow 3 x-2 y=0
\end{aligned}\)
Slope of line \(A B=\frac{2}{2}=1\)
Given, \(\mathrm{OE} \perp \mathrm{AB}\)
\(\therefore\) Slope of \(\mathrm{OE}=-1\)
Equation of OE is \(y=m x\)
\(\begin{aligned}
& \Rightarrow y=-x \\
& \Rightarrow x+y=0
\end{aligned}\)
\(\therefore\) Joint equation of median and altitude is
\(\begin{aligned}
& (3 x-2 y)(x+y)=0 \\
& \Rightarrow 3 x^2+x y-2 y^2=0
\end{aligned}\)
\(\begin{aligned}
\therefore \quad & D \equiv\left(\frac{1+3}{2}, \frac{2+4}{2}\right) \\
& \Rightarrow D \equiv(2,3)
\end{aligned}\)

Equation of OD is \(y=m x\)
\(\begin{aligned}
& \Rightarrow y=\frac{3}{2} x \\
& \Rightarrow 3 x-2 y=0
\end{aligned}\)
Slope of line \(A B=\frac{2}{2}=1\)
Given, \(\mathrm{OE} \perp \mathrm{AB}\)
\(\therefore\) Slope of \(\mathrm{OE}=-1\)
Equation of OE is \(y=m x\)
\(\begin{aligned}
& \Rightarrow y=-x \\
& \Rightarrow x+y=0
\end{aligned}\)
\(\therefore\) Joint equation of median and altitude is
\(\begin{aligned}
& (3 x-2 y)(x+y)=0 \\
& \Rightarrow 3 x^2+x y-2 y^2=0
\end{aligned}\)
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