The ends of the base of an isosceles triangle | vedclass

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(D) Let the vertices of the triangle be $C(2a, 0)$,$B(0, a)$,and $A(2a, k)$.Since the triangle is isosceles with base $BC$,we have $AB = AC$.The lengt

Vedclass · Accepted Answer

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

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(D) Let the vertices of the triangle be $C(2a, 0)$,$B(0, a)$,and $A(2a, k)$.Since the triangle is isosceles with base $BC$,we have $AB = AC$.The length $AC = |k - 0| = |k|$.The length $AB = \sqrt{(2a - 0)^2 + (k - a)^2} = \sqrt{4a^2 + (k - a)^2}$.Equating $AB^2 = AC^2$,we get $4a^2 + (k - a)^2 = k^2$.$4a^2 + k^2 - 2ak + a^2 = k^2$.$5a^2 = 2ak$.Since $a 
eq 0$,we have $k = \frac{5a}{2}$.Thus,the vertex $A$ is $(2a, \frac{5a}{2})$.The other side is the line passing through $B(0, a)$ and $A(2a, \frac{5a}{2})$.The slope $m = \frac{\frac{5a}{2} - a}{2a - 0} = \frac{\frac{3a}{2}}{2a} = \frac{3}{4}$.The equation of the line is $y - a = \frac{3}{4}(x - 0)$.$4y - 4a = 3x$,which simplifies to $3x - 4y + 4a = 0$.

The ends of the base of an isosceles triangle are at $(2a, 0)$ and $(0, a)$. The equation of one side is $x = 2a$. The equation of the other side is

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