Question

Find the coordinates of the circumcenter of triangle PQR with vertices P(-2,5) Q(4,1) and R(-2,-3) This is what I’m struggling with the most. I have found all the midpoints and the slopes.

Answer 1

In this type of question, for us to be able to get the coordinates of the triangle's circumcenter, the following are the steps that we are going to do.

Step 1: Determine the coordinates of at least two midpoints of the triangle.

Step 2: Generate the formula of the 2 bisecting lines (from vertex to midpoint)

Step 3: Equating the equations of two bisecting lines to find their point of intersection which is also the triangle's circumcenter.

We get,

Step 1: Determine the coordinates of at least two midpoints of the triangle.

Midpoint between P and Q,

`\text{ Midpoint of P and Q = }\frac{\text{ -2 + 4}}{\text{ 2}},\text{ }\frac{\text{ 5 + 1}}{\text{ 2 }}`
`\text{ = 1, 3}`

Midpoint between Q and R,

`\text{ Midpoint between Q and R = }\frac{\text{ 4 + (-2)}}{2},\text{ }\frac{\text{ 1 + (-3)}}{2}`
`\text{ = 1, }-1`

Step 2: Generate the formula of the 2 bisecting lines (from vertex to midpoint).

Equation 1: Midpoint P & Q to Vertex R.

`\text{ Slope = m= }\frac{\text{ -3 - 3}}{-2\text{ - 1}}\text{ = }\frac{\text{ -6}}{-3}\text{ = 2}`
`\begin{gathered} \text{ y = mx + b} \\ \text{ -3 = (2)(-2) + b} \\ -3\text{ = -4 + b} \\ \text{ b = -3 + 4} \\ \text{ b = 1 (y-intercept, b)} \end{gathered}`

The equation is, therefore,

`\text{ y = mx + b}`
`\text{ y = (2)x +(1)}`
`\text{ y = 2x }+\text{ 1}`

Equation 2: Midpoint Q and R to vertex P.

`\text{ Slope = m = }\frac{5\text{ - (-1)}}{-2\text{ - 1}}\text{ = }\frac{\text{ 6}}{\text{ -3}}\text{ = -2}`
`\begin{gathered} \text{ y = mx + b} \\ \text{ 5 = -2(-2) }+\text{ b} \\ \text{ 5 = 4 + b} \\ \text{ b = 5 - 4} \\ \text{ b = 1 (y-intercept)} \end{gathered}`

The equation is, therefore,

`\text{ y = mx + b}`
`\text{ y = -2(x) + }1`
`\text{ y = -2x + 1}`

Step 3: Equating the equations of two bisecting lines to find their point of intersection which is also the triangle's circumcenter.

Using the substitution method,

y = 2x + 1

y = -2x + 1

2x + 1 = -2x + 1

2x + 2x = 1 - 1

4x = 0

4x/4 = 0/4

x = 0

At x = 0,

y = 2(0) + 1

y = 1

Therefore, the coordinate of the circumcenter of the triangle PQR is 0, 1

Plotting the triangle will be,