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?

1 Answer
May 26, 2016

I have taken you nearly to the end. It just a matter of algebraic manipulation.
#color(blue)("Someone else may be able to think of a simpler method!")#
#color(blue)("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.
#color(brown)("It also gets you extra marks")#

Tony B
Given:
Length of side B is equal to length of side C
Coordinates of vertex AB#->(x,y)->(3,2)#
Coordinates of vertex AC#->(x,y)->(5,2)#
Area of triangle is #16color(white)(.)"units"^2#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Method:")#

Step 1. Relate area to length of side A to determine height h
Step 2. Determine point #P_1 -> (x_c,y_c)#
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->(x_v,y_v)#

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

#color(green)(|A| = sqrt( (5-3)^2+(2-1)^2) = sqrt(4+1)=sqrt(5))#
#color(brown)( |A|" means the magnitude of A")#

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

#=>16=sqrt(5)/2xxh" "=>" "color(green)(h=32/sqrt(5))#
'................................................................
#color(blue)(("Step 2")#

Point #P_1# is the mean of the two ends of side A

#color(green)(=>P_1->(x_c,y_c) = ((5+3)/2,(2+1)/2)->(4,3/2))#
'................................................................
#color(blue)(("Step 3")#

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

#color(green)(=> m= (1-2)/(5-3) = -1/2)#
'................................................................
#color(blue)(("Step 4")#
Standard form equation: #y=mx+c#

In this case the gradient is #-1/m ->-1/[-1/2] =+2#

#y=2x+c# which passes through point #P_1->(x_c,y_c)=(4,3/2)#

Substitution gives #3/2=2(4)+c" "=>" "c= 3/2-8 = -13/2#

#color(green)(=> y= 2x-13/2)#
'..............................................................
#color(blue)(("Step 5")#

#h=" distance " P_1 to P_2#

#=>h^2=(x_v-x_c)^2+(y_v-y_c)^2#

#(32/sqrt(5))^2=(x_v-4)^2+(y_v-3/2)^2#

But from step 4 #y_v=2x_v-13/2# so by substitution

#(32/sqrt(5))^2=[x_v-4]^2+[(2x_v-13/2)-3/2]^2#

'.....................................................................
#color(magenta)("I will let you take it from this point")#

#color(brown)("Solve for "x_v" then substitute in " y_v= 2x_v-13/2)#

#color(green)("Don't forget that the vertex below the base is also a possible answer!")#