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.
[Math] 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
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Best Answer
Rearranging the equations,
$$L_1:y=\frac12x+2$$ $$L_2:y=-2x+5$$
As product of slope of $L_1$ and $L_2$ $=(-2)\cdot(\frac12)=-1$, $L_1$ is perpendicular to $L_2$.
Let length of one of the side of triangle be $r$.
$$\frac12r^2=10$$ $$r=\sqrt{20}$$
Let a point on $L_1$ be A: $(t,\frac{t+4}2)$, by considering the distance between the intersection point $(\frac65,\frac{13}5)$ and A,we have
$$\sqrt{20}=\sqrt{(t-\frac65)^2+(\frac{t+4}2-\frac{13}5)^2}$$ $$t=5.2\text{ or }-2.8$$
$$A_1= (5.2,4.6)\text{ or }A_2=(-2.8,0.6)$$
Similarly, for a point on $L_2$, B: $(p,-2p+5)$, by considering the distance between the intersection point $(\frac65,\frac{13}5)$ and B,we have
$$\sqrt{20}=\sqrt{(p-\frac65)^2+(-2p+5-\frac{13}5)^2}$$ $$p=3.2\text{ or }-0.8$$ $$B_1= (3.2,-1.4)\text{ or }B_2=(-0.8,6.6)$$
So we have line $L_{11}=A_1B_1, L_{21}=A_2B_1,L_{22}=A_2B_2,L_{12}=A_1B_2$
$$L_{11}:\frac{y-4.6}{x-5.2}=\frac{4.6-(-1.4)}{5.-3.2}$$ $$L_{11}:y=3x-11$$
Similar for $L_{21},L_{22}\text{ and }L_{12}$.