JEE MAIN - Mathematics (2024 - 29th January Evening Shift - No. 2)
Let $$\mathrm{A}$$ be the point of intersection of the lines $$3 x+2 y=14,5 x-y=6$$ and $$\mathrm{B}$$ 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
$$\frac{13}{2}$$
8
$$\frac{5}{2}$$
6
Explanation
Solving lines $$\mathrm{L}_1(3 \mathrm{x}+2 \mathrm{y}=14)$$ and $$\mathrm{L}_2(5 \mathrm{x}-\mathrm{y}=6)$$ to get $$\mathrm{A}(2,4)$$ and solving lines $$\mathrm{L}_3(4 \mathrm{x}+3 \mathrm{y}=8)$$ and $$\mathrm{L}_4(6 \mathrm{x}+\mathrm{y}=5)$$ to get $$\mathrm{B}\left(\frac{1}{2}, 2\right)$$.
Finding Eqn. of $$\mathrm{AB}: 4 \mathrm{x}-3 \mathrm{y}+4=0$$
Calculate distance PM
$$\Rightarrow\left|\frac{4(5)-3(-2)+4}{5}\right|=6$$
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