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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.

Find the coordinates of the circumcenter of triangle PQR with vertices P(-2,5) Q(4,1) and-example-1
User Fuenfundachtzig
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1 Answer

16 votes
16 votes

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,

Find the coordinates of the circumcenter of triangle PQR with vertices P(-2,5) Q(4,1) and-example-1
User Pars
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3.0k points