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JEE Mains · Maths · STD 11 - 9. straight line
Let \(A (-3, 2)\) and \(B (-2, 1)\) be the vertices of a triangle \(ABC\). If the centroid of this triangle lies on the line \(3x + 4y + 2 = 0\), then the vertex \(C\) lies on the line
- A \(4x +3y+5= 0\)
- B \(3x +4y+3=0\)
- C \(4x +3y+3=0\)
- D \(3x + 4y + 5 = 0\)
Answer & Solution
Correct Answer
(B) \(3x +4y+3=0\)
Step-by-step Solution
Detailed explanation
Let \(C\left( {{x_1},{y_1}} \right)\) Centroid, \(E = \left( {\frac{{{x_1} - 5}}{3},\frac{{{x_1} - 3}}{3}} \right)\) Since centroid lies on the line \(3x + 4y + 2 = 0\) \(\therefore 3\left( {\frac{{{x_1} - 5}}{3}} \right) + 4\left( {\frac{{{x_1} - 3}}{3}} \right) + 2 = 0\)…
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