# An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from (5 ,1 ) to (3 ,2 ) and the triangle's area is 12 , what are the possible coordinates of the triangle's third corner?

Jul 8, 2016

Two possibilities: $\left(8.8 , 11.1\right)$ or $\left(- 0.8 , - 8.1\right)$

#### Explanation:

Refer to the figure below It's known that:
${P}_{2} \left(5 , 1\right)$, ${P}_{3} \left(3 , 2\right)$
$S = 12$
$B = C$
It's asked ${P}_{1}$

$A = \sqrt{{\left(5 - 3\right)}^{2} + {\left(2 - 1\right)}^{2}} = \sqrt{5}$
$S = \frac{\text{base"*"height}}{2} \implies 12 = \frac{\sqrt{5} \cdot h}{2} \implies h = \frac{24}{\sqrt{5}}$
${B}^{2} = {\left(\frac{A}{2}\right)}^{2} + {h}^{2} = {\left(\frac{\sqrt{5}}{2}\right)}^{2} + {\left(\frac{24}{\sqrt{5}}\right)}^{2} = \frac{5}{4} + \frac{576}{5} = \frac{2329}{20} \implies B \cong 10.791$ (this is the distance ${P}_{1} {P}_{2}$ or ${P}_{1} {P}_{3}$)

Now we could use the equations of the distance between 2 points to determine ${P}_{1}$, but since ${P}_{1}$ is directly above (or below) $M \left(4 , 1.5\right)$ we can find ${P}_{1}$ in the following easier way:

${k}_{{P}_{2} {P}_{3}} = \frac{\Delta y}{\Delta x} = \frac{1 - 2}{5 - 3} = - \frac{1}{2}$
line ${P}_{2} {P}_{3}$: $\left(y - 1\right) = \left(- \frac{1}{2}\right) \left(x - 5\right) \implies 2 y - 2 = - x + 5 \implies x + 2 y - 7 = 0$ 
Distance between a point and a line
$d = | a {x}_{0} + b {y}_{0} + c \frac{|}{\sqrt{{a}^{2} + {b}^{2}}}$
So, making $d = h$
$\frac{24}{\cancel{\sqrt{5}}} = | x + 2 y - 7 \frac{|}{\cancel{\sqrt{5}}}$
We get two equations:
(I) $24 = x + 2 y - 7 \implies 2 y = - x + 31 \implies y = - \frac{x}{2} + 15.5$
(II) $- 24 = x + 2 y - 7 \implies 2 y = - x - 17 \implies y = - \frac{x}{2} - 8.5$

Since ${P}_{1}$ is also in the bisector line of side A, let's find its equation
$p = - \frac{1}{k} _ \left({P}_{2} {P}_{3}\right) = 2$
line ${P}_{1} M$: $\left(y - 1.5\right) = 2 \left(x - 4\right) \implies y = 2 x - 8 + 1.5 \implies y = 2 x - 6.5$

Finding the two possible answers (by combining  and  or  and ):

(I) $2 x - 6.5 = - \frac{x}{2} + 15.5 \implies 2.5 x = 22 \implies x = 8.8$
$\to y = 2 \cdot 8.8 - 6.5 \implies y = 11.1$
$\implies \left(8.8 , 11.1\right)$

(II) $2 x - 6.5 = - \frac{x}{2} - 8.5 \implies 2.5 x = - 2 \implies x = - 0.8$
$\to y = 2 \cdot \left(- 0.8\right) - 6.5 \implies y = - 8.1$
$\implies \left(- 0.8 , - 8.1\right)$

Checking the results:
Notice that $\frac{8.8 - 0.8}{2} = 4 = {x}_{M}$
and $\frac{11.1 - 8.1}{2} = 1.5 = {y}_{M}$
Notice also that
(testing result I)${P}_{1} {P}_{2} = \sqrt{{3.8}^{2} + {10.1}^{2}} = \sqrt{116.45} \cong 10.791$
(testing result I)${P}_{1} {P}_{3} = \sqrt{{5.8}^{2} + {9.1}^{2}} = \sqrt{116.45} \cong 10.791$
(Try result II)
As it should be: all checked