Question : A line passing through the origin perpendicularly cuts the line $3x–2y=6$ at point M. Find the co-ordinates of M.
Option 1: $(\frac{18}{13},\frac{12}{13})$
Option 2: $(\frac{18}{13},-\frac{12}{13})$
Option 3: $(-\frac{18}{13},-\frac{12}{13})$
Option 4: $(-\frac{18}{13},\frac{12}{13})$
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Correct Answer: $(\frac{18}{13},-\frac{12}{13})$
Solution :
Given: The line $3x–2y=6$. -------------------------------(1)
Comparing this to $y=mx+c$, we have,
⇒ $2y=3x–6$
⇒ $y=\frac{3x–6}{2}$
⇒ $y=\frac{3x}{2}–3$
$m=\frac{3}{2}$ and $c=–3$
Since the lines are perpendicular to each other,
$m_1×m_2=–1$
⇒ $\frac{3}{2}×m_2=–1$
⇒ $m_2=\frac{–2}{3}$
The line is passing through the origin. So, $c=0$
⇒ $y=mx$
⇒ $y=\frac{–2}{3}×x$
⇒ $3y=–2x$
⇒ $2x+3y=0$ ---------------------------------------------(2)
From equations (1) and (2), we have
$9x–6y=18$
$4x+6y=0$
⇒ $13x=18$
⇒ $x=\frac{18}{13}$ and $y=\frac{–12}{13}$
Hence, the correct answer is $(\frac{18}{13},\frac{–12}{13})$.
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