An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from #(7 ,1 )# to #(8 ,5 )# and the triangle's area is #27 #, what are the possible coordinates of the triangle's third corner?

Question

An isosceles triangle has sides A, B, and C with sides B and C being equal in length.  If side A goes from #(7  ,1  )# to #(8  ,5 )# and the triangle's area is #27  #, what are the possible coordinates of the triangle's third corner?

Caroline Aucoin · Accepted Answer

#(6.55882, 3.23529) and (8.44118, 2.76471)#

Let's begin by finding the length of side "a":

#a = sqrt((8-7)^2 + (5 - 1)^2)#

#a = sqrt(1^2 + 4^2)#

#a = sqrt(17)#

Use side "a" as the base of the triangle, find the height:

#Area = 1/2base xx height#

#2 = 1/2sqrt(17) xx height#

#height = (4sqrt(17))/17#

The height must bisect side "a" at the point #(7.5, 3)#

We can used the distance formula to write one equation for the two possible points:

#(4sqrt(17))/17 = sqrt((x - 7.5)^2 + (y - 3)^2)#

Square both sides:

#16/17 = (x - 7.5)^2 + (y - 3)^2" [1]"#

We need one more equation. Find the slope of side "a":

#m = (5 - 1)/(8 - 7) = 4#

Because the height is perpendicular to side "a", its slope is: #-1/4#

We can use the slope and the point to find the equation of line for the height:

#3 = -1/4(7.5) + b#

#b = 39/8#

The equation for the height is:

#y = -1/4x + 39/8" [2]"#

Solving equations [1] and [2] is very long and tedious process. I gave them to WolframAlpha. Here are the points:

#(6.55882, 3.23529) and (8.44118, 2.76471)#

Allison Allen · Answer

undefined Given the coordinates of two vertices of the triangle, you can find the length of side $A$ using the distance formula. Then, you can use the formula for the area of a triangle to find the height corresponding to side $A$. Once you have the height, you can determine the possible coordinates of the third corner by considering the fact that it lies on a line parallel to side $A$ and at a distance equal to the height from the line segment connecting the given two vertices.

Here are the steps:

1. Calculate the length of side $A$ using the distance formula: 
   $$ A = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$
   Substitute the coordinates: $ (7,1) $ and $ (8,5) $.

2. Use the formula for the area of a triangle:
   $$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $$
   Substitute the known values: $\text{Area} = 27$ and $\text{base} = A$.

3. Solve for the height:
   $$ 27 = \frac{1}{2} \times A \times \text{height} $$
   Solve for $\text{height}$.

4. Determine the possible coordinates of the third corner:
   Since the triangle is isosceles, the third corner lies on a line parallel to side $A$ and at a distance equal to the height from the line segment connecting the given two vertices. You can find the equation of the line parallel to side $A$ passing through one of the given vertices and then use the distance and direction to find the coordinates of the third corner.

5. Repeat step 4 for both given vertices to find the possible coordinates of the third corner.