Question Statement

Find the equations of the right bisectors and the point of concurrency of the triangle $XYZ$ when $X(0,0)$, $Y(7,0)$ and $Z(7,4)$. What is the name of the point of concurrency?

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Background and Explanation

A right bisector (or perpendicular bisector) of a side of a triangle is a line perpendicular to that side passing through its midpoint. The three right bisectors of a triangle always meet at a single point. This problem requires finding midpoints, slopes of sides, perpendicular slopes, and equations of lines.

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Solution

Finding the Midpoints

Let $M_1$, $M_2$ and $M_3$ be the midpoints of sides $\overline{XY}$, $\overline{YZ}$ and $\overline{ZX}$ respectively.

Using the midpoint formula $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$:

$M_1=\left(\frac{0+7}{2},\frac{0+0}{2}\right)=\left(\frac{7}{2},0\right)$

$M_2=\left(\frac{7+7}{2},\frac{4+0}{2}\right)=(7,2)$

$M_3=\left(\frac{7+0}{2},\frac{4+0}{2}\right)=\left(\frac{7}{2},2\right)$

Finding Slopes of Sides

$\text{Slope of side }\overline{XY}=\frac{0-0}{7-0}=\frac{0}{7}=0$

$\text{Slope of side }\overline{YZ}=\frac{4-0}{7-7}=\frac{4}{0}=\infty$

$\text{Slope of side }\overline{ZX}=\frac{4-0}{7-0}=\frac{4}{7}$

Right Bisector of Side $\overline{XY}$ (Line $\overline{M_{1}A}$)

The right bisector $\overline{M_{1}A}$ is perpendicular to side $\overline{XY}$. For perpendicular lines, the slope is the negative reciprocal:

$\text{Slope of }\overline{M_{1}A}=\frac{-1}{\text{Slope of side }\overline{XY}}=\frac{-1}{0}=\infty$

Since the slope is infinite (vertical line), the right bisector is a vertical line passing through $M_1\left(\frac{7}{2},0\right)$. Using point-slope form:

$y-0=\infty \left(x-\frac{7}{2}\right)$

$\Rightarrow \frac{y}{\left(x-\frac{7}{2}\right)}=\infty$

This is possible only when the denominator equals zero:

$x-\frac{7}{2}=0$

$\Rightarrow 2x-7=0$

Right Bisector of Side $\overline{YZ}$ (Line $\overline{M_{2}B}$)

The right bisector $\overline{M_{2}B}$ is perpendicular to side $\overline{YZ}$:

$\text{Slope of }\overline{M_{2}B}=\frac{-1}{\text{Slope of side }\overline{YZ}}=\frac{-1}{\infty }=0$

Since the slope is 0 (horizontal line), using point-slope form through $M_2(7,2)$:

$y-2=0(x-7)$

$\Rightarrow y-2=0$

Right Bisector of Side $\overline{ZX}$ (Line $\overline{M_{3}C}$)

The right bisector $\overline{M_{3}C}$ is perpendicular to side $\overline{ZX}$:

$\text{Slope of }\overline{M_{3}C}=\frac{-1}{\text{Slope of side }\overline{ZX}}=\frac{-1}{4/7}=\frac{-7}{4}$

Using point-slope form through $M_3\left(\frac{7}{2},2\right)$ with slope $\frac{-7}{4}$:

$y-2=\frac{-7}{4}\left(x-\frac{7}{2}\right)$

$y-2=\frac{-7}{4}x+\frac{49}{8}$

Multiply both sides by 8:

$8y-16=-14x+49$

$\Rightarrow 14x+8y-65=0$

Finding the Point of Concurrency

To find where all three right bisectors meet, solve any two equations simultaneously. Using equations (i) and (ii):

From (i): $2x=7\Rightarrow x=\frac{7}{2}$

From (ii): $y=2$

Thus, $\left(\frac{7}{2},2\right)$ is the point of concurrency of these right bisectors.

Name of the Point of Concurrency

The point of concurrency of the right bisectors of a triangle is called the circumcentre. This point is equidistant from all three vertices of the triangle and is the center of the circumscribed circle (circumcircle).

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Key Formulas or Methods Used

• Midpoint formula: $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$
• Slope formula: $m=\frac{y_2-y_1}{x_2-x_1}$
• Perpendicular slope: $m_{\perp }=\frac{-1}{m}$ (negative reciprocal)
• Point-slope form: $y-y_1=m(x-x_1)$
• Vertical line: $x=a$ (when slope is undefined)
• Horizontal line: $y=b$ (when slope is zero)

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Summary of Steps

1. Find midpoints of all three sides using the midpoint formula
2. Calculate slopes of all three sides of the triangle
3. Find perpendicular slopes for each right bisector using negative reciprocals
4. Write equations of right bisectors using point-slope form through each midpoint
5. Solve any two equations simultaneously to find the point of concurrency
6. Identify the point of concurrency as the circumcentre