JEE Mains · Maths · STD 11 - 9. straight line
The equations of two sides \(\mathrm{AB}\) and \(\mathrm{AC}\) of a triangle \(\mathrm{ABC}\) are \(4 \mathrm{x}+\mathrm{y}=14\) and \(3 \mathrm{x}-2 \mathrm{y}=5\), respectively. The point \(\left(2,-\frac{4}{3}\right)\) divides the third side \(\mathrm{BC}\) internally in the ratio \(2: 1\). The equation of the side \(\mathrm{BC}\) is :
- A \(x-6 y-10=0\)
- B \(x-3 y-6=0\)
- C \(x+3 y+2=0\)
- D \(x+6 y+6=0\)
Answer & Solution
Correct Answer
(C) \(x+3 y+2=0\)
Step-by-step Solution
Detailed explanation
\( \frac{2 x_2+x_1}{3}=2, \frac{2\left(\frac{3 x_2-5}{2}\right)+\left(14-4 x_1\right)}{3}=\frac{-4}{3} \) \( 2 x_2+x_1=6,3 x_2-4 x_1=-13 \) \( x_2=1, x_1=4\) So, \(\mathrm{C}(1,-1), \mathrm{B}(4,-2)\) \(\mathrm{m}=\frac{-1}{3}\) Equation of \(B C: y+1=\frac{-1}{3}(x-1)\)…
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