# An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from (2 ,5 ) to (8 ,7 ) and the triangle's area is 12 , what are the possible coordinates of the triangle's third corner?

Nov 17, 2016

The possible points are: $\left(3.8 , 9.6\right)$ and $\left(6.2 , 2.4\right)$

#### Explanation:

The length of side "a" is:

$a = \sqrt{{\left(8 - 2\right)}^{2} + {\left(7 - 5\right)}^{2}} = \sqrt{40} = 2 \sqrt{10}$

The area of a triangle is:

$\text{Area" = (1/2)"Base"xx"Height}$

Using side "a" as the base of the triangle and the given area, we can compute the height:

$12 = \left(\frac{1}{2}\right) 2 \sqrt{10} \times \text{Height}$

$\text{Height} = \frac{12}{\sqrt{10}}$

Think of the height as the radius of a circle upon which the two possible points must lie:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {\left(\frac{12}{\sqrt{10}}\right)}^{2}$

Because the triangle's other two sides are the same length, the center of this circle must be the midpoint between the points #(2, 5) and (8, 7):

$h = 2 + \frac{8 - 2}{2} = 5 \mathmr{and} k = 5 + \frac{7 - 5}{2} = 6$

Substitute into the equation of the circle:

${\left(x - 5\right)}^{2} + {\left(y - 6\right)}^{2} = {\left(\frac{12}{\sqrt{10}}\right)}^{2} \text{ [1]}$

The two points must, also, lie on a line that is perpendicular to side "a"; the slope, m, of this line is:

$m = \frac{2 - 8}{7 - 5} = - \frac{6}{2} = - 3$

Using, this slope, the center point, and the point-slope form of the equation of a line, we write the equation upon which the two points must lie:

$y = - 3 \left(x - 5\right) + 6 \text{ [2]}$

Here is a graph of the equation [1], equation [2] and the two given points:

Substitute the right side of equation [2] into equation [1]:

${\left(x - 5\right)}^{2} + {\left(- 3 \left(x - 5\right) + 6 - 6\right)}^{2} = {\left(\frac{12}{\sqrt{10}}\right)}^{2}$

${\left(x - 5\right)}^{2} + {\left(- 3 \left(x - 5\right)\right)}^{2} = {\left(\frac{12}{\sqrt{10}}\right)}^{2}$

${\left(x - 5\right)}^{2} + 9 {\left(x - 5\right)}^{2} = {\left(\frac{12}{\sqrt{10}}\right)}^{2}$

${\left(x - 5\right)}^{2} + 9 {\left(x - 5\right)}^{2} = \frac{144}{10}$

$10 {\left(x - 5\right)}^{2} = \frac{144}{10}$

${\left(x - 5\right)}^{2} = \frac{144}{100}$

$x - 5 = \pm \frac{12}{10}$

$x = 5 \pm \frac{12}{10}$

$x = 3.8 \mathmr{and} x = 6.2$

To obtain the corresponding y values, substitute these x values into equation [2]

$y = - 3 \left(3.8 - 5\right) + 6$ and $y = - 3 \left(6.2 - 5\right) + 6$

$y = 9.6 \mathmr{and} y = 2.4$