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Two equal sides of an isosceles triangle...

Two equal sides of an isosceles triangle are given by `7x-y+3=0` and `x+y=3`, and its third side passes through the point `(1,-10)`. Find the equation of the third side.

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Let AP and AQ be the internal and external bisectors of `angleABC.` Since `AB=AC` therefore `AP botBC and AQ||BC as AP bot AQ,` Thus BC passes through `R(1,-10)` and is parallel to AQ and RC' passes throug `R(1,-10)` and is parallel to internal bgisectors AP.

The equation of the two bisector of angle BAC are
`(7x-y+3)/(sqrt50)=+-(x+y-3)/(sqrt2)`
`implies3x+y-3=0and x -3y+9=0.` The line through `(1,-10)` parallel to `3x+y-3=0` is `3x+y-k=0` where `3-10-k=0,k=-7,`
i.e., the strainght line through `(1,-10)` and parallel to `3x+y-3=0` is given by `x-3y-31=0.`
Hence the third side in its two positions (BC and B'C') have the equations `3x+y+7=0and x-3y-31=0.`
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