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Question

Two equal sides of an isosceles triangle are given by the equations 7xy+3=0 and x+y3=0 and its third side passes through the point (1,10). Determine the equation of the third side.

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Solution

Since, Δ is isosceles triangle so, third side is equally inclined to the lines 7xy+3=0 and x+y3=0

Hence, third line is parallel to their angle in sector.

7xy+350=±x+y32

3x+y3=0 -----(i)

x3y+9=0 -----(ii)


The line through (1,10) parallel to (i) is 3x+y+7=0 and that parallel to (ii) is x3y31=0.

Which is the required answer.

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