JEE Mains · Maths · STD 11 - 9. straight line
Let the mid points of the sides of a triangle \(ABC\) be \(\left(\dfrac{5}{2}, 7\right)\), \(\left(\dfrac{5}{2}, 3\right)\) and \((4, 5)\). If its incentre is \((h, k)\), then \(3h + k\) is equal to :
- A \(11\)
- B \(12\)
- C \(13\)
- D \(14\)
Answer & Solution
Correct Answer
(C) \(13\)
Step-by-step Solution
Detailed explanation
Let the vertices of the triangle \(ABC\) be \(A(x_1, y_1)\), \(B(x_2, y_2)\), and \(C(x_3, y_3)\). Let the given midpoints be of sides \(BC\), \(CA\), and \(AB\) respectively: Midpoint of \(BC = \left(\dfrac{5}{2}, 7\right) \Rightarrow x_2 + x_3 = 5\) and \(y_2 + y_3 = 14\)…
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