Question:
Let \(A(-1, 1)\), \(B(3, 4)\) and \(C(2, 0)\) be given three points.
A line \(y = mx\), \(m > 0\), intersects lines AC and BC at point P and Q respectively. Let \(A_1\) and \(A_2\) be the areas of \(\Delta ABC\) and \(\Delta PQC\) respectively, such that \(A_1 = 3A_2\), then the value of \(m\) is equal to :
Let \(A(-1, 1)\), \(B(3, 4)\) and \(C(2, 0)\) be given three points.
A line \(y = mx\), \(m > 0\), intersects lines AC and BC at point P and Q respectively. Let \(A_1\) and \(A_2\) be the areas of \(\Delta ABC\) and \(\Delta PQC\) respectively, such that \(A_1 = 3A_2\), then the value of \(m\) is equal to :
A line \(y = mx\), \(m > 0\), intersects lines AC and BC at point P and Q respectively. Let \(A_1\) and \(A_2\) be the areas of \(\Delta ABC\) and \(\Delta PQC\) respectively, such that \(A_1 = 3A_2\), then the value of \(m\) is equal to :
Show Hint
Find P and Q in terms of m; both triangles have the vertex C, so areas follow from cross products.
Updated On: Nov 4, 2026
- 1
- 3
- 2
- \(\frac{4}{15}\)
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The Correct Option is A
Solution and Explanation
Line AC: slope \(\frac{0 - 1}{2 + 1} = -\frac{1}{3}\), \(y = -\frac{x-2}{3}\). With \(y = mx\): \(P = \left(\frac{2}{3m+1},\frac{2m}{3m+1}\right)\).
Line BC: slope \(\frac{4-0}{3-2} = 4\), \(y = 4(x-2)\). With \(y = mx\): \(Q = \left(\frac{8}{4-m},\frac{8m}{4-m}\right)\).
\(\overrightarrow{CP} = \frac{2m}{3m+1}(-3, 1)\), \(\overrightarrow{CQ} = \frac{2m}{4-m}(1, 4)\).
\(A_2 = \frac{1}{2}\left|\overrightarrow{CP}\times\overrightarrow{CQ}\right| = \frac{1}{2}\cdot\frac{4m^2}{(3m+1)|4-m|}\cdot|(-3)(4) - (1)(1)| = \frac{13}{2}\cdot\frac{4m^2}{(3m+1)|4-m|}\).
\(A_1 = \frac{1}{2}\left|(-1)(4-0) + 3(0-1) + 2(1-4)\right| = \frac{13}{2}\).
\(A_1 = 3A_2\Rightarrow (3m+1)|4-m| = 12m^2\).
For \(0<m<4\): \((3m+1)(4-m) = 12m^2\Rightarrow 15m^2 - 11m - 4 = 0\Rightarrow m = 1\) or \(-\frac{4}{15}\); only \(m = 1\) is positive. For \(m>4\): \(9m^2 + 11m + 4 = 0\) has no real roots.
So \(m = 1\).
Answer: (1) 1
Line BC: slope \(\frac{4-0}{3-2} = 4\), \(y = 4(x-2)\). With \(y = mx\): \(Q = \left(\frac{8}{4-m},\frac{8m}{4-m}\right)\).
\(\overrightarrow{CP} = \frac{2m}{3m+1}(-3, 1)\), \(\overrightarrow{CQ} = \frac{2m}{4-m}(1, 4)\).
\(A_2 = \frac{1}{2}\left|\overrightarrow{CP}\times\overrightarrow{CQ}\right| = \frac{1}{2}\cdot\frac{4m^2}{(3m+1)|4-m|}\cdot|(-3)(4) - (1)(1)| = \frac{13}{2}\cdot\frac{4m^2}{(3m+1)|4-m|}\).
\(A_1 = \frac{1}{2}\left|(-1)(4-0) + 3(0-1) + 2(1-4)\right| = \frac{13}{2}\).
\(A_1 = 3A_2\Rightarrow (3m+1)|4-m| = 12m^2\).
For \(0<m<4\): \((3m+1)(4-m) = 12m^2\Rightarrow 15m^2 - 11m - 4 = 0\Rightarrow m = 1\) or \(-\frac{4}{15}\); only \(m = 1\) is positive. For \(m>4\): \(9m^2 + 11m + 4 = 0\) has no real roots.
So \(m = 1\).
Answer: (1) 1
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