Question

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?

Answer 1

Coordinates of third vertex A of equilateral triangle ABC is

`color(blue)((3.125, 5.875))`

Explanation:

Length of side BC = `a = sqrt((5-1)^2 + (8-4)^2) = 5.6569`

Height `h = (2 * A_t) / a = (2*15)/5.6569 = 5.3033`

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

`b = sqrt((5.6569/2)^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 = (8-4)/(5-1) = 1`

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

Equation of AD is
`y - 6 = -1 * (x - 3)`

`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(x - 1)`

`y - 0.8824x = 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

`color(blue)((3.125, 5.875))`

Answer 2

`color(magenta)(6.9,2.1`

Explanation:

`sqrt((x_2-x_1)^2+(y_2-y_1))`

`:.=sqrt((1-5)^2+(4-8)^2)`

`:.=sqrt((-4)^2+(-4)^2)`

`:.=sqrt(16+16)`

`:.=sqrt32`

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

`:.Area triangle=1/2xxbasexxh=15`

`:.0.5xx5.656854249xxh=15`

`:.2.828427125xxh=15`

`:.h=15/2.828427125`

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

`:.In triangle ABC:- 1/2 base=2.828427125`

`:.5.503300858/2.828427125=tanB=1.945710678=62^@ 47'57''`

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

`:.180^@-(62^@ 47' 57''+62^@ 47' 57'')=54^@ 24' 06''=angle A`

Bearing `A-B=342^@ 12' 03''-180^@= 162^@ 12' 03''=`BearingB-A

`:.162^@ 12' 03''-54^@ 24' 06''=107^@ 47' 57''=`Bearing A-C

Bearing A-B`= 342^@ 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=y`coord A

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

~~~~~~~~~~~~~~~~~~~~~~~~~~~

Check:-

Bearing `A-C=107^@ 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