An isosceles triangle has sides $a, b$ $a, b$ and $c$ $c$ with sides $b$ $b$ and $c$ $c$ being equal in length. If side $a$ $a$ goes from $(2, 9)$ $(2 ,9 )$ to $(8, 5)$ $(8 ,5 )$ and the triangle's area is $64$ $64$ , what are the possible coordinates of the triangle's third corner?

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An isosceles triangle has sides $a, b$ $a, b$ and $c$ $c$ with sides $b$ $b$ and $c$ $c$ being equal in length. If side $a$ $a$ goes from $(2, 9)$ $(2 ,9 )$ to $(8, 5)$ $(8 ,5 )$ and the triangle's area is $64$ $64$ , what are the possible coordinates of the triangle's third corner?

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Answer 1 · 2018-06-18T12:25:55

$(- \frac{63}{13}, - \frac{101}{13}) or (\frac{193}{13}, \frac{283}{13})$ $(-63/13, -101/13) or (193/13, 283/13)$

Explanation:

Side $a$ $a$ is $(2, 9)$ $(2,9)$ to $(8, 5)$ $(8,5)$

Let's do this directly. We have

$a^{2} = {(8 - 2)}^{2} + {(5 - 9)}^{2} = 36 + 16 = 52$ $a^2 =(8-2)^2+(5-9)^2=36+16=52$

Area $S = \frac{1}{2} a h$ $S = 1/2 a h$ or

$h = \frac{2 S}{a} = \frac{2 (64)}{\sqrt{52}} = \frac{128}{\sqrt{52}}$ $h = {2S}/a = {2(64)}/\sqrt{52} = 128/sqrt{52}$

The other vertex is along the perpendicular bisector of side $a$ $a$ . The direction vector for side as is $(8 - 2, 5 - 9) = (6, - 4)$ $(8-2,5-9)=(6,-4)$ so the perpendicular through the midpoint is

$(x, y) = (\frac{2 + 8}{2}, \frac{9 + 5}{2}) + t (4, 6) = (5, 7) + t (4, 6)$ $(x,y) = ( {2+8}/2, {9+5}/2) + t(4,6) = (5,7)+t(4,6)$

We need $| t (4, 6) | = h$ $|t(4,6)|=h$

$t = \pm \frac{h}{| (4.6) |} = \pm \frac{\frac{128}{\sqrt{52}}}{\sqrt{52}} = \pm \frac{128}{52} = \pm \frac{32}{13}$ $t = pm h/|(4.6)| = pm (128/sqrt{52})/sqrt{52} = pm 128/52 = pm 32/13$

Two choices for the third vertex:

$(x, y) = (5, 7) \pm (\frac{32}{13}) (4, 6) = (- \frac{63}{13}, - \frac{101}{13}) or (\frac{193}{13}, \frac{283}{13})$ $(x,y) = (5,7) pm (32/13) (4,6) = (-63/13, -101/13) or (193/13, 283/13)$

Let's check one:

$b^{2} = {(- \frac{63}{13} - 2)}^{2} + {(- \frac{101}{13} - 9)}^{2} = \frac{4265}{13}$ $b^2 = (-63/13 - 2)^2 + (-101/13 - 9)^2 = 4265/13$

$c^2 = (-63/13 - 8)^2 + (-101/13 - 5)^2 = 4265/13 quad sqrt$

Area $S$

$16S^2 = 4(52)(4265/13) - (52)^2$

$S = 64 quad sqrt$

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An isosceles triangle has sides $a, b$ $a, b$ and $c$ $c$ with sides $b$ $b$ and $c$ $c$ being equal in length. If side $a$ $a$ goes from $(2, 9)$ $(2 ,9 )$ to $(8, 5)$ $(8 ,5 )$ and the triangle's area is $64$ $64$ , what are the possible coordinates of the triangle's third corner?

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Explanation:

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An isosceles triangle has sides a, ba,ba, b and ccc with sides bbb and ccc being equal in length. If side aaa goes from (2 ,9 )(2,9)(2 ,9 ) to (8 ,5 )(8,5)(8 ,5 ) and the triangle's area is 64 6464 , what are the possible coordinates of the triangle's third corner?

1 Answer

Explanation:

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An isosceles triangle has sides $a, b$ $a, b$ and $c$ $c$ with sides $b$ $b$ and $c$ $c$ being equal in length. If side $a$ $a$ goes from $(2, 9)$ $(2 ,9 )$ to $(8, 5)$ $(8 ,5 )$ and the triangle's area is $64$ $64$ , what are the possible coordinates of the triangle's third corner?