JEE Mains · Maths · STD 11 - 9. straight line
Two sides of a rhombus are along the lines, \(x -y+ 1 = 0\) and \(7x-y-5 =0.\) If its diagonals intersect at \((-1,-2),\) then which one of the following is a vertex of this rhombus?
- A \(\left( {\frac{1}{3}, - \frac{8}{3}} \right)\)
- B \(\left( { - \frac{{10}}{3}, - \frac{7}{3}} \right)\)
- C \(\left( { - 3, - 9} \right)\)
- D \(\;\left( { - 3, - 8} \right)\)
Answer & Solution
Correct Answer
(A) \(\left( {\frac{1}{3}, - \frac{8}{3}} \right)\)
Step-by-step Solution
Detailed explanation
Equation of angle bisector of the lines \(x-y+1=0\) and \(7 \mathrm{x}-\mathrm{y}-5=0\) is given by \(\frac{x-y+1}{\sqrt{2}}=\pm \frac{7 x-y-5}{5 \sqrt{2}}\) \(\Rightarrow 5(\mathrm{x}-\mathrm{y}+1)=7 \mathrm{x}-\mathrm{y}-5\) and \(5(x-y+1)=-7 x+y+5\)…
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