JEE Mains · Maths · STD 11 - 9. straight line
Let \(A\) be the point of intersection of the lines \(3 x+\) \(2 y=14,5 x-y=6\) and \(B\) be the point of intersection of the lines \(4 x+3 y=8,6 x+y=5\) The distance of the point \(P(5,-2)\) from the line \(\mathrm{AB}\) is
- A \(\frac{13}{2}\)
- B \(8\)
- C \(\frac{5}{2}\)
- D \(6\)
Answer & Solution
Correct Answer
(D) \(6\)
Step-by-step Solution
Detailed explanation
Solving lines \(L_1(3 x+2 y=14)\) and \(L_2(5 x-y=6)\) to get \(A(2,4)\) and solving lines \(L_3(4 x+3 y=8)\) and \(L_4(6 x+y=5)\) to get \(B\left(\frac{1}{2}, 2\right)\). Finding Eqn. of \(A B: 4 x-3 y+4=0\) Calculate distance \(PM\)…
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