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?

2 Answers
Jan 12, 2018

Coordinates of third vertex A of equilateral triangle ABC is

#color(blue)((3.125, 5.875))#

Explanation:

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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))#

Jan 14, 2018

#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