Question:
Let C be the centroid of the triangle with vertices (3, \(-1\)), (1, 3) and (2, 4). Let P be the point of intersection of the lines \(x+3y-1=0\) and \(3x-y+1=0\). Then the line passing through the points C and P also passes through the point :
Let C be the centroid of the triangle with vertices (3, \(-1\)), (1, 3) and (2, 4). Let P be the point of intersection of the lines \(x+3y-1=0\) and \(3x-y+1=0\). Then the line passing through the points C and P also passes through the point :
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Find C and P, form the line CP, and test the four points.
Updated On: Nov 5, 2026
- \((-9,-7)\)
- \((9,7)\)
- \((7,6)\)
- \((-9,-6)\)
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The Correct Option is D
Solution and Explanation
\(C=\left(\frac{3+1+2}{3},\frac{-1+3+4}{3}\right)=(2,2)\).
Solving \(x+3y=1\) and \(3x-y=-1\): from the second \(y=3x+1\), so \(x+9x+3=1\Rightarrow x=-\frac15\), \(y=\frac25\); \(P=\left(-\frac15,\frac25\right)\).
Slope of CP \(=\frac{2-\frac25}{2+\frac15}=\frac{8/5}{11/5}=\frac8{11}\).
Line: \(y-2=\frac8{11}(x-2)\Rightarrow 8x-11y+6=0\).
Test: \((-9,-6)\): \(-72+66+6=0\) (lies on it). \((-9,-7)\) gives 11, \((9,7)\) gives 1, \((7,6)\) gives -4.
Answer: (4) \((-9,-6)\)
Solving \(x+3y=1\) and \(3x-y=-1\): from the second \(y=3x+1\), so \(x+9x+3=1\Rightarrow x=-\frac15\), \(y=\frac25\); \(P=\left(-\frac15,\frac25\right)\).
Slope of CP \(=\frac{2-\frac25}{2+\frac15}=\frac{8/5}{11/5}=\frac8{11}\).
Line: \(y-2=\frac8{11}(x-2)\Rightarrow 8x-11y+6=0\).
Test: \((-9,-6)\): \(-72+66+6=0\) (lies on it). \((-9,-7)\) gives 11, \((9,7)\) gives 1, \((7,6)\) gives -4.
Answer: (4) \((-9,-6)\)
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