Question:
Points \((h,k)\), \((1,2)\), \((-3,4)\) lie on \(L_1\). Line \(L_2\) through \((h,k)\) and \((4,3)\) is perpendicular to \(L_1\). Find \(\frac{k}{h}\).
Points \((h,k)\), \((1,2)\), \((-3,4)\) lie on \(L_1\). Line \(L_2\) through \((h,k)\) and \((4,3)\) is perpendicular to \(L_1\). Find \(\frac{k}{h}\).
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Use \(m_1m_2=-1\) whenever two lines are perpendicular.
Updated On: Jun 9, 2026
- \(\frac14\)
- \(\frac13\)
- \(-\frac17\)
- \(-\frac15\)
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The Correct Option is B
Solution and Explanation
Step 1: Find equation of \(L_1\).
Slope of \(L_1\):
\[
m_1=\frac{4-2}{-3-1}
=
-\frac12
\]
Equation:
\[
x+2y-5=0
\]
Since \((h,k)\) lies on \(L_1\),
\[
h+2k=5
\]
Step 2: Use perpendicularity. \[ m_2=2 \] Hence \[ \frac{3-k}{4-h}=2 \] \[ 2h-k=5 \]
Step 3: Solve simultaneously. \[ h+2k=5 \] \[ 2h-k=5 \] Solving, \[ h=3,\qquad k=1 \] Thus \[ \frac{k}{h} = \frac13 \] \[ \boxed{\frac13} \]
Step 2: Use perpendicularity. \[ m_2=2 \] Hence \[ \frac{3-k}{4-h}=2 \] \[ 2h-k=5 \]
Step 3: Solve simultaneously. \[ h+2k=5 \] \[ 2h-k=5 \] Solving, \[ h=3,\qquad k=1 \] Thus \[ \frac{k}{h} = \frac13 \] \[ \boxed{\frac13} \]
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