Question

find the point on the straight line 3x+y+4=0 which is equidistant from the points (-5,6) and (3,2)

Answer 1

Dear student
$$\begin{gathered} 3\mathrm{x}+\mathrm{y}+4=0-5,63,2 \\ \mathrm{Let}\mathrm{the}\mathrm{point}\mathrm{be}\mathrm{P}(\mathrm{x},\mathrm{y}) \\ \mathrm{Then}3\mathrm{x}+\mathrm{y}+4=0....\left(1\right) \\ \mathrm{Since}\mathrm{the}\mathrm{line}\left(1\right)\mathrm{is}\mathrm{equidistant}\mathrm{from}(-5,6)\mathrm{and}(3,2) \\ \mathrm{So},(\mathrm{x}+5{)}^2+(\mathrm{y}-6{)}^2=(\mathrm{x}-3{)}^2+(\mathrm{y}-2{)}^2 \\ \Rightarrow {\mathrm{x}}^2+25+25\mathrm{x}+{\mathrm{y}}^2+36-12\mathrm{y}={\mathrm{x}}^2+9-6\mathrm{x}+{\mathrm{y}}^2+4-4\mathrm{y} \\ \Rightarrow 31\mathrm{x}-8\mathrm{y}+48=0...\left(2\right) \\ \mathrm{Solving}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ \mathrm{x}=-\frac{16}{11}\mathrm{and}\mathrm{y}=\frac{4}{11} \\ \mathrm{So},\mathrm{P}=\left(-\frac{16}{11},\frac{4}{11}\right) \end{gathered}$$
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