In an isosceles triangle,the ends of its base | vedclass

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(C) Let the vertices be $A(0, a)$,$B(2a, 0)$,and $C(x_1, y_1)$.Since one of the sides is a horizontal line and it is not the $X$-axis,the side $AC$ mu

Vedclass · Accepted Answer

$\frac{9a}{4}$

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(C) Let the vertices be $A(0, a)$,$B(2a, 0)$,and $C(x_1, y_1)$.Since one of the sides is a horizontal line and it is not the $X$-axis,the side $AC$ must be horizontal.Therefore,the $y$-coordinate of $C$ must be equal to the $y$-coordinate of $A$,so $y_1 = a$.Since the triangle is isosceles and $AC$ is a side,we have $AC = BC$ or $AC = AB$ or $BC = AB$.Given $AC$ is horizontal,$AC = \sqrt{(x_1 - 0)^2 + (a - a)^2} = |x_1|$.$BC = \sqrt{(x_1 - 2a)^2 + (a - 0)^2} = \sqrt{(x_1 - 2a)^2 + a^2}$.Setting $AC^2 = BC^2$,we get $x_1^2 = (x_1 - 2a)^2 + a^2$.$x_1^2 = x_1^2 - 4ax_1 + 4a^2 + a^2$.$4ax_1 = 5a^2$.Since $a 
eq 0$,$x_1 = \frac{5a}{4}$.Thus,$x_1 + y_1 = \frac{5a}{4} + a = \frac{9a}{4}$.

In an isosceles triangle,the ends of its base are $(2a, 0)$ and $(0, a)$. One of its two other sides is a horizontal line (not the $X$-axis). If the third vertex is $(x_1, y_1)$,then $x_1 + y_1 =$

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