JEE Mains · Maths · STD 11 - 9. straight line
In a triangle \(ABC\), coordianates of \(A\) are \((1, 2)\) and the equations of the medians through \(B\) and \(C\) are \(x + y = 5\) and \(x = 4\) respectively. Then area of \(\Delta ABC\) (in sq. units) is
- A \(5\)
- B \(9\)
- C \(12\)
- D \(4\)
Answer & Solution
Correct Answer
(B) \(9\)
Step-by-step Solution
Detailed explanation
Median through \(C\) is \(x=4\) So the \(x\) coordianate of \(C\) is \(4\). let \(C \equiv \left( {4,y} \right)\) then the midpoint of \(A(1,2)\) and \(C(4,y)\) is \(D\) which lies on the median through \(B\). \(D \equiv \left( {\frac{{1 + 4}}{2},\frac{{2 + y}}{2}} \right)\)…
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