# An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from (1 ,4 ) to (5 ,8 ) and the triangle's area is 15 , what are the possible coordinates of the triangle's third corner?

Jan 12, 2018

Coordinates of third vertex A of equilateral triangle ABC is

$\textcolor{b l u e}{\left(3.125 , 5.875\right)}$

#### Explanation:

Length of side BC = $a = \sqrt{{\left(5 - 1\right)}^{2} + {\left(8 - 4\right)}^{2}} = 5.6569$

Height $h = \frac{2 \cdot {A}_{t}}{a} = \frac{2 \cdot 15}{5.6569} = 5.3033$

Length of side b = c = sqrt((a/2)^2 + h^2)

$b = \sqrt{{\left(\frac{5.6569}{2}\right)}^{2} + {5.3033}^{2}} = 6.01$

Coordinates of center point of BC = D is ((5+1)/2, (8+4)/2) = (3,6)

Slope of line segment BC $= m = \frac{8 - 4}{5 - 1} = 1$

Slope of altitude AD = -(1/m) = -1

$y - 6 = - 1 \cdot \left(x - 3\right)$

$x + y = 9$ Eqn (1)

Slope of line BA = m1 = h / b = 5.3033 / 6.01 = 0.8824

Equation of line BA is

$y - 4 = 0.8824 \left(x - 1\right)$

$y - 0.8824 x = 3.1176$ Eqn (2)

Solving Eqns (1), (2) we get the coordinates of the third vertex A.

$x = 3.125 , y = 5.875$

Coordinates of third vertex A of equilateral triangle ABC is

$\textcolor{b l u e}{\left(3.125 , 5.875\right)}$

Jan 14, 2018

color(magenta)(6.9,2.1

#### Explanation:

$\sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + \left({y}_{2} - {y}_{1}\right)}$

$\therefore = \sqrt{{\left(1 - 5\right)}^{2} + {\left(4 - 8\right)}^{2}}$

$\therefore = \sqrt{{\left(- 4\right)}^{2} + {\left(- 4\right)}^{2}}$

$\therefore = \sqrt{16 + 16}$

$\therefore = \sqrt{32}$

color(magenta):.color(magenta)(=5.656854249= Distance B-C

$\therefore A r e a \triangle = \frac{1}{2} \times b a s e \times h = 15$

$\therefore 0.5 \times 5.656854249 \times h = 15$

$\therefore 2.828427125 \times h = 15$

$\therefore h = \frac{15}{2.828427125}$

color(magenta):.color(magenta)(h=5.303300858

$\therefore I n \triangle A B C : - \frac{1}{2} b a s e = 2.828427125$

$\therefore \frac{5.503300858}{2.828427125} = \tan B = 1.945710678 = {62}^{\circ} 47 ' 57 ' '$

62^@ 47'57''-45^@=72^@ 12' 03''+270^@=342^2 12' 03''=Bearing B-A

$\therefore {180}^{\circ} - \left({62}^{\circ} 47 ' 57 ' ' + {62}^{\circ} 47 ' 57 ' '\right) = {54}^{\circ} 24 ' 06 ' ' = \angle A$

Bearing $A - B = {342}^{\circ} 12 ' 03 ' ' - {180}^{\circ} = {162}^{\circ} 12 ' 03 ' ' =$BearingB-A

$\therefore {162}^{\circ} 12 ' 03 ' ' - {54}^{\circ} 24 ' 06 ' ' = {107}^{\circ} 47 ' 57 ' ' =$Bearing A-C

Bearing A-B$= {342}^{\circ} 12 ' 03 ' '$ at distance $6.187617789$

$\cos = 0.952133838 \sin = - 0.305681456$

 6.187617789xx0.952133838 =5.891440278+1color(magenta)(=6.891440278=x coord A

 6.187617789xx-0.305681456=-1.89144002+4color(magenta)(=2.10855998=ycoord A

color(magenta)(:.A=6.9,2.1

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Check:-

Bearing $A - C = {107}^{\circ} 47 ' 57 ' '$Distance 6.187617789

cos 107^@ 47' 57''=-0.305681456xx6.187617789=-1.89144002color(magenta)(+6.891440278color(magenta)(=5.000000258=x  of C

sin107^@ 47' 57''=0.952133838xx6.187617789=5.891440278color(magenta)(+2.10855998color(magenta)(=8.000000258=y of C