The base of an isosceles triangle has its end | vedclass

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(D) Let the vertices of the base be $P(2a, 0)$ and $Q(0, a)$.Since one side is parallel to the $y$-axis,the third vertex $R$ must have the same $x$-co

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$3x - 4y + 4a = 0$

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(D) Let the vertices of the base be $P(2a, 0)$ and $Q(0, a)$.Since one side is parallel to the $y$-axis,the third vertex $R$ must have the same $x$-coordinate as either $P$ or $Q$.Case $1$: If $R$ has $x$-coordinate $0$,then $R = (0, y)$. Since the triangle is isosceles,$RP = RQ$.$RP^2 = (2a - 0)^2 + (0 - y)^2 = 4a^2 + y^2$.$RQ^2 = (0 - 0)^2 + (a - y)^2 = (a - y)^2$.$4a^2 + y^2 = a^2 - 2ay + y^2 \implies 2ay = -3a^2 \implies y = -\frac{3a}{2}$.The line $PR$ passes through $(2a, 0)$ and $(0, -\frac{3a}{2})$.The slope $m = \frac{-3a/2 - 0}{0 - 2a} = \frac{-3a/2}{-2a} = \frac{3}{4}$.The equation is $y - 0 = \frac{3}{4}(x - 2a) \implies 4y = 3x - 6a \implies 3x - 4y - 6a = 0$.Case $2$: If $R$ has $x$-coordinate $2a$,then $R = (2a, y)$.$RQ^2 = (2a - 0)^2 + (y - a)^2 = 4a^2 + (y - a)^2$.$RP^2 = (2a - 2a)^2 + (y - 0)^2 = y^2$.$4a^2 + y^2 - 2ay + a^2 = y^2 \implies 5a^2 = 2ay \implies y = \frac{5a}{2}$.The line $QR$ passes through $(0, a)$ and $(2a, \frac{5a}{2})$.The slope $m = \frac{5a/2 - a}{2a - 0} = \frac{3a/2}{2a} = \frac{3}{4}$.The equation is $y - a = \frac{3}{4}(x - 0) \implies 4y - 4a = 3x \implies 3x - 4y + 4a = 0$.

The base of an isosceles triangle has its endpoints at $(2a, 0)$ and $(0, a)$. One side is parallel to the $y$-axis. Find the equation of the other side.

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