# An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from (7 ,4 ) to (2 ,1 ) and the triangle's area is 15 , what are the possible coordinates of the triangle's third corner?

Jul 22, 2017

The coordinates of the third corner are $\left(7.14 , - 1.9\right)$ or $\left(1.85 , 6.92\right)$

#### Explanation:

The length of the base of the isoceles triangle is

$b = \sqrt{{\left(7 - 2\right)}^{2} + {\left(4 - 1\right)}^{2}} = \sqrt{25 + 9} = \sqrt{34}$

The area of the triangle is

$a r e a = \frac{1}{2} b h = 15$

The altitude is $= \frac{30}{b} = \frac{30}{\sqrt{34}}$

The mid-point of the base is

$= \left(\frac{7 + 2}{2} , \frac{4 + 1}{2}\right) = \left(\frac{9}{2} , \frac{5}{2}\right)$

The slope of the base is $m = \frac{1 - 4}{2 - 7} = - \frac{3}{-} 5 = \frac{3}{5}$

The slope of the altitude is $m ' = - \frac{1}{m} = - \frac{5}{3}$

The equation of the altitude is

$y - \frac{5}{2} = - \frac{5}{3} \left(x - \frac{9}{2}\right)$

$y = - \frac{5}{3} x + \frac{15}{2} + \frac{5}{2} = - \frac{5}{3} x + 10$........................$\left(1\right)$

Let the coordinates of the third corner be $= \left(x , y\right)$

Then,

${\left(x - \frac{9}{2}\right)}^{2} + {\left(y - \frac{5}{2}\right)}^{2} = {\left(\frac{30}{\sqrt{34}}\right)}^{2}$

${\left(x - \frac{9}{2}\right)}^{2} + {\left(y - \frac{5}{2}\right)}^{2} = \frac{900}{34} = 26.5$..................$\left(2\right)$

Solving for $x$ and $y$ in equations $\left(1\right)$ and $\left(2\right)$

${\left(x - 4.5\right)}^{2} + {\left(- \frac{5}{3} x + 10 - 2.5\right)}^{2} = 26.5$

${\left(x - 4.5\right)}^{2} + {\left(- \frac{5}{3} x + 7.5\right)}^{2} = 26.5$

${x}^{2} - 9 x + 20.25 + 2.78 {x}^{2} - 25 x + 56.25 - 26.5 = 0$

$3.78 {x}^{2} - 34 x + 50 = 0$

$x = \frac{34 \pm \sqrt{{34}^{2} - 4 \cdot 3.78 \cdot 50}}{2 \cdot 3.78}$
$x = \frac{34 \pm \sqrt{400}}{7.56} = \frac{34 \pm 20}{7.56}$
${x}_{1} = 7.14$ or ${x}_{2} = 1.85$
${y}_{1} = - \frac{5}{3} \cdot 7.14 + 10 = - 1.9$
${y}_{2} = - \frac{5}{3} \cdot 1.85 + 10 = 6.92$