The vertices of the base of an isosceles tria | vedclass

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(C) Given the parabola $y^2=4x$ and the line $y=2x-4$. To find the vertices of the base,substitute $y=2x-4$ into $y^2=4x$: $(2x-4)^2 = 4x$ $4(x-2)^2 =

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$\left(\frac{9}{2}, 0\right)$

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(C) Given the parabola $y^2=4x$ and the line $y=2x-4$. To find the vertices of the base,substitute $y=2x-4$ into $y^2=4x$: $(2x-4)^2 = 4x$ $4(x-2)^2 = 4x$ $x^2-4x+4 = x$ $x^2-5x+4 = 0$ $(x-1)(x-4) = 0$ So,$x=1$ or $x=4$. For $x=1$,$y=2(1)-4 = -2$. Point $C = (1, -2)$. For $x=4$,$y=2(4)-4 = 4$. Point $B = (4, 4)$. Let the third vertex be $A = (x, 0)$ on the $X$-axis. Since the triangle is isosceles with base $BC$,we have $AB=AC$,so $AB^2 = AC^2$. $(x-4)^2 + (0-4)^2 = (x-1)^2 + (0-(-2))^2$ $x^2-8x+16+16 = x^2-2x+1+4$ $-8x+32 = -2x+5$ $6x = 27$ $x = \frac{27}{6} = \frac{9}{2}$. Thus,the coordinates of the third vertex are $\left(\frac{9}{2}, 0\right)$.

The vertices of the base of an isosceles triangle lie on a parabola $y^2=4x$ and the base is a part of the line $y=2x-4$. If the third vertex of the triangle lies on the $X$-axis,its coordinates are

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