# 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 16 , what are the possible coordinates of the triangle's third corner?

May 26, 2016

I have taken you nearly to the end. It just a matter of algebraic manipulation.
$\textcolor{b l u e}{\text{Someone else may be able to think of a simpler method!}}$
$\textcolor{b l u e}{\text{Perhaps utilising transformation}}$

#### Explanation:

In geometry it always a good idea to draw a quick sketch. It helps clarify what you need to do and also shows the person marking your work that you have given thought about the solution.
$\textcolor{b r o w n}{\text{It also gets you extra marks}}$

Given:
Length of side B is equal to length of side C
Coordinates of vertex AB$\to \left(x , y\right) \to \left(3 , 2\right)$
Coordinates of vertex AC$\to \left(x , y\right) \to \left(5 , 2\right)$
Area of triangle is $16 \textcolor{w h i t e}{.} {\text{units}}^{2}$
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$\textcolor{b l u e}{\text{Method:}}$

Step 1. Relate area to length of side A to determine height h
Step 2. Determine point ${P}_{1} \to \left({x}_{c} , {y}_{c}\right)$
Step 3. Determine gradient of line for side A

Step 4. Determine equation of line ${P}_{1} {P}_{2}$
Step 5. Using h determine ${P}_{2} \to \left({x}_{v} , {y}_{v}\right)$

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color(blue)(("Step 1")
Using Pythagoras determine length side A

$\textcolor{g r e e n}{| A | = \sqrt{{\left(5 - 3\right)}^{2} + {\left(2 - 1\right)}^{2}} = \sqrt{4 + 1} = \sqrt{5}}$
$\textcolor{b r o w n}{| A | \text{ means the magnitude of A}}$

"Area" = ("base")/2xxh

$\implies 16 = \frac{\sqrt{5}}{2} \times h \text{ "=>" } \textcolor{g r e e n}{h = \frac{32}{\sqrt{5}}}$
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color(blue)(("Step 2")

Point ${P}_{1}$ is the mean of the two ends of side A

$\textcolor{g r e e n}{\implies {P}_{1} \to \left({x}_{c} , {y}_{c}\right) = \left(\frac{5 + 3}{2} , \frac{2 + 1}{2}\right) \to \left(4 , \frac{3}{2}\right)}$
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color(blue)(("Step 3")

gradient (m) = ("change in y")/("change in x")" "  reading left to right

$\textcolor{g r e e n}{\implies m = \frac{1 - 2}{5 - 3} = - \frac{1}{2}}$
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color(blue)(("Step 4")
Standard form equation: $y = m x + c$

In this case the gradient is $- \frac{1}{m} \to - \frac{1}{- \frac{1}{2}} = + 2$

$y = 2 x + c$ which passes through point ${P}_{1} \to \left({x}_{c} , {y}_{c}\right) = \left(4 , \frac{3}{2}\right)$

Substitution gives $\frac{3}{2} = 2 \left(4\right) + c \text{ "=>" } c = \frac{3}{2} - 8 = - \frac{13}{2}$

$\textcolor{g r e e n}{\implies y = 2 x - \frac{13}{2}}$
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color(blue)(("Step 5")

$h = \text{ distance } {P}_{1} \to {P}_{2}$

$\implies {h}^{2} = {\left({x}_{v} - {x}_{c}\right)}^{2} + {\left({y}_{v} - {y}_{c}\right)}^{2}$

${\left(\frac{32}{\sqrt{5}}\right)}^{2} = {\left({x}_{v} - 4\right)}^{2} + {\left({y}_{v} - \frac{3}{2}\right)}^{2}$

But from step 4 ${y}_{v} = 2 {x}_{v} - \frac{13}{2}$ so by substitution

${\left(\frac{32}{\sqrt{5}}\right)}^{2} = {\left[{x}_{v} - 4\right]}^{2} + {\left[\left(2 {x}_{v} - \frac{13}{2}\right) - \frac{3}{2}\right]}^{2}$

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$\textcolor{m a \ge n t a}{\text{I will let you take it from this point}}$

$\textcolor{b r o w n}{\text{Solve for "x_v" then substitute in } {y}_{v} = 2 {x}_{v} - \frac{13}{2}}$

$\textcolor{g r e e n}{\text{Don't forget that the vertex below the base is also a possible answer!}}$