Question:
In a triangle ABC, coordinates of A are \((1, 2)\) and the equations of the medians through B and C are respectively, \(x + y = 5\) and \(x = 4\). Then area of \(\Delta\) ABC (in sq. units) is :
In a triangle ABC, coordinates of A are \((1, 2)\) and the equations of the medians through B and C are respectively, \(x + y = 5\) and \(x = 4\). Then area of \(\Delta\) ABC (in sq. units) is :
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The centroid is the intersection of the two medians; use it to find B and C, then apply the area formula.
Updated On: Nov 11, 2026
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- 9
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The Correct Option is D
Solution and Explanation
Centroid G is the intersection of \(x+y=5\) and \(x=4\): \(G=(4,1)\).
B lies on \(x+y=5\): let \(B=(b,5-b)\). C lies on \(x=4\): let \(C=(4,c)\).
Centroid: \(\frac{1+b+4}{3}=4\Rightarrow b=7\); \(\frac{2+(5-b)+c}{3}=1\Rightarrow 7-7+c=3\Rightarrow c=3\).
So \(B=(7,-2)\), \(C=(4,3)\).
Area \(=\frac{1}{2}|1(-2-3)+7(3-2)+4(2+2)|=\frac{1}{2}|-5+7+16|=\frac{18}{2}=9\).
Answer: (4) 9
B lies on \(x+y=5\): let \(B=(b,5-b)\). C lies on \(x=4\): let \(C=(4,c)\).
Centroid: \(\frac{1+b+4}{3}=4\Rightarrow b=7\); \(\frac{2+(5-b)+c}{3}=1\Rightarrow 7-7+c=3\Rightarrow c=3\).
So \(B=(7,-2)\), \(C=(4,3)\).
Area \(=\frac{1}{2}|1(-2-3)+7(3-2)+4(2+2)|=\frac{1}{2}|-5+7+16|=\frac{18}{2}=9\).
Answer: (4) 9
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