<s>[INST] If $x-2y+4=0$ and $2x+y-5=0$ are the sides of isosceles triangle having area $10$ sq unit .Equation of third side is? Okay, I know two sides of an  isosceles triangle are equal . I have also taken out the intersection points of the lines given in the question . Other than this , I have no clue about how I will find out the equation of the third side of the isosceles triangle . Please help.
 [/INST] The lines: $x-2y+4=0$ & $2x+y-5=0$ are normal to each other thus the triangle is an isosceles right triangle. Thus you are looking for the third side representing the hypotenuse of isosceles right triangle. Let $a$ be length of two equal & perpendicular sides of isosceles right triangle then the area of triangle is given as $$\frac{1}{2}(a)(a)=10 \implies a=\sqrt{20}=2\sqrt{5}$$ The equations of lines bisecting the angle (right angle here) between the given lines: $x-2y+4=0$ & $2x+y-5=0$ are given as $$\frac{x-2y+4}{\sqrt{1^2+(-2)^2}}=\pm \frac{2x+y-5}{\sqrt{2^2+1^2}}\implies x-2y+4=\pm(2x+y-5)$$$$ \implies x+3y-9=0 \quad \text{&} \quad 3x-y-1=0 $$ 
Case 1: Let the equation of third (unknown) line (representing the hypotenuse of isosceles right triangle) be $3x-y+c=0$ normal to the first right-angle bisector: $x+3y-9=0$. Now, solving: $3x-y+c=0$ & any given line say $2x+y-5=0$, we get the intersection point $\left(\frac{5-c}{5}, \frac{2c+15}{5} \right)$. Similarly, solving both the equations of the given lines, we get intersection point $\left(\frac{6}{5}, \frac{13}{5} \right)$. Now, calculating the length $a=2\sqrt{5}$ of equal sides of isosceles triangle using distance formula as
$$\sqrt{\left(\frac{6}{5}-\frac{5-c}{5}\right)^2+\left(\frac{13}{5}-\frac{2c+15}{5}\right)^2}=a=2\sqrt{5}$$$$ c^2+2c-99=0 \implies c=9 \quad \text{&} \quad c=-11$$ Thus by setting the values of $c$, we get two equations: $3x-y+9=0$ & $3x-y-11=0$ lying on either side of right-angled vertex of given isosceles (right) triangle.        
Case 2: Let the equation of third (unknown) line (representing the hypotenuse of isosceles right triangle) be $x+3y+c=0$ normal to the second right-angle bisector: $3x-y-1=0$. Now, solving: $x+3y+c=0$ & $2x+y-5=0$, we get the intersection point $\left(\frac{c+15}{5}, \frac{-(2c+5)}{5} \right)$. Similarly, solving both the equations of the given lines, we get intersection point $\left(\frac{6}{5}, \frac{13}{5} \right)$. Now, calculating the length $a=2\sqrt{5}$ of equal sides of isosceles triangle using distance formula as
$$\sqrt{\left(\frac{6}{5}-\frac{c+15}{5}\right)^2+\left(\frac{13}{5}-\frac{-(2c+5)}{5}\right)^2}=a=2\sqrt{5}$$$$ c^2+18c-19=0 \implies c=1 \quad \text{&} \quad c=-19$$ Thus by setting the values of $c$, we get two equations: $x+3y+1=0$ & $x+3y-19=0$ lying on either side of right-angled vertex of given isosceles (right) triangle.          
Thus, there are total four lines: $3x-y-11=0$, $3x-y+9=0$, $x+3y+1=0$ & $x+3y-19=0$ representing the third unknown side of the isosceles (right) triangle satisfying all the conditions provided in the question. Hence, you may select any of them as your choice.  
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