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Perpendicular Bisector

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A **perpendicular bisector** is a line segment that:

- intersects another line segment at a right angle (90
^{o}), and - divides the intersected line segment into two equal parts.

The point of intersection of the perpendicular bisector with a line segment is the **midpoint** of the line segment.

The diagram below shows a graphical representation of a perpendicular bisector crossing a line segment on a Cartesian plane.

The perpendicular bisector crosses the midpoint of the points A (x_{1}, y_{1}) and B (x_{2}, y_{2}) that lie on the line segment. This is denoted by the coordinates M (x_{m}, y_{m}). The distance from the midpoint to either point A or B are of equal length. In other words, AM = BM.

Let the equation of the line containing the points A and B be y = m_{1 }x + c where m_{1} is the slope of that line. Similarly, let the equation of the perpendicular bisector of this line be y = m_{2 }x + d where m_{2} is the slope of the perpendicular bisector.

The slope of a line can also be referred to as the gradient.

As the two lines, y = m_{1 }x + c and y = m_{2 }x + d are perpendicular to each other, the product between the two slopes m_{1} and m_{2} is -1.

${m}_{1}\times {m}_{2}=-1$

Referring back to the diagram above, say we are given the coordinates of two points A (x_{1}, y_{1}) and B (x_{2}, y_{2}). We want to find the equation of the perpendicular bisector that crosses the midpoint between A and B. We can locate the equation of the perpendicular bisector using the following method.

**Step 1:** Given points A (x_{1}, y_{1}) and B (x_{2}, y_{2}), find the coordinates of the midpoint using the Midpoint Formula.

$M({x}_{m},{y}_{m})=\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)$

**Step 2:** Calculate the slope of the line segment, m_{1}, connecting A and B using the Gradient Formula.

${m}_{1}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$

**Step 3:** Determine the slope of the perpendicular bisector, m_{2}, using the derivation below.

${m}_{1}\times {m}_{2}=-1\Rightarrow {m}_{2}=-\frac{1}{{m}_{1}}$

**Step 4:** Evaluate the equation of the perpendicular bisector using the Equation of a Line Formula and the found midpoint M (x_{m}, y_{m}) and slope m_{2}.

$y-{y}_{m}={m}_{2}(x-{x}_{m})$

Find the equation of the perpendicular bisector of the line segment joining the points (9, -3) and (-7, 1).

**Solution**

Let (x_{1}, y_{1}) = (9, -3) and (x_{2}, y_{2}) = (-7, 1).

The midpoint is given by:

$({x}_{m},{y}_{m})=\left(\frac{9+(-7)}{2},\frac{-3+1}{2}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow ({x}_{m},{y}_{m})=\left(\frac{2}{2},-\frac{2}{2}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow ({x}_{m},{y}_{m})=\left(1,-1\right)$

The slope of the line segment joining the points (9, -3) and (-7, 1) is:

${m}_{1}=\frac{1-(-3)}{-7-9}\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{1}=-\frac{4}{16}\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{1}=-\frac{1}{4}$

The slope of the perpendicular bisector of this line segment is:

$-\frac{1}{4}{m}_{2}=-1\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{m}_{2}}{4}=1\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{2}=4$

We thus obtain the equation of the perpendicular bisector as:

$y-(-1)=4(x-1)\phantom{\rule{0ex}{0ex}}\Rightarrow y+1=4x-4\phantom{\rule{0ex}{0ex}}\Rightarrow y=4x-4-1\phantom{\rule{0ex}{0ex}}\Rightarrow y=4x-5$

The Perpendicular Bisector Theorem tells us that any point on the perpendicular bisector is equidistant from both the endpoints of a line segment.

A point is said to be **equidistant **from a set of coordinates if the distances between that point and each coordinate in the set are equal.

Observe the diagram below.

If the line MO is the perpendicular bisector of the line XY then:

$\overline{)XM}=\overline{)YM}\phantom{\rule{0ex}{0ex}}\overline{)XA}=\overline{)YA}\phantom{\rule{0ex}{0ex}}\overline{)XB}=\overline{)YB}\phantom{\rule{0ex}{0ex}}\overline{)XC}=\overline{)YC}$

**Proof**

Before we begin the proof, recall the SAS Congruence rule.

**SAS Congruence**

If two sides and an included angle of one triangle are equal to two sides and an included angle of another triangle then the triangles are congruent.

Observe the sketch above. Comparing triangles XAM and YAM we find that:

XM = YM since M is the midpoint

AM = AM because it is a shared side

∠XMA = ∠YMA = 90

^{o }

By the SAS Congruence rule, triangles XAM and YAM are congruent. Using CPCTC, A is equidistant from both X and Y, or in other words, XA = YA as corresponding parts of congruent triangles.

Given the triangle XYZ below, determine the length of the side XZ if the perpendicular bisector of the line segment BZ is XA for the triangle XBZ. Here, XB = 17 cm and AZ = 6 cm.

Since AX is the perpendicular bisector of the line segment BZ, any point on AX is equidistant from points B and Z by the Perpendicular Bisector Theorem. This implies that XB = XZ. Thus XZ = 17 cm.

The Converse of the Perpendicular Bisector Theorem states that if a point is equidistant from the endpoints of a line segment in the same plane, then that point lies on the perpendicular bisector of the line segment.

To get a clearer picture of this, refer to the sketch below.

If XP = YP then the point P lies on the perpendicular bisector of the line segment XY.

**Proof**

Observe the diagram below.

We are given that XA = YA. We want to prove that XM = YM. Construct a perpendicular line from point A that intersects the line XY at point M. This forms two triangles, XAM and YAM. Comparing these triangles, notice that

XA = YA (given)

AM = AM (shared side)

∠XMA = ∠YMA = 90

^{o}

By the SAS Congruence rule, triangles XAM and YAM are congruent. As point A is equidistant from both X and Y then A lies on the perpendicular bisector of the line XY. Thus, XM = YM, and M is equidistant from both X and Y as well.

Given the triangle XYZ below, determine the length of the sides AY and AZ if XZ = XY = 5 cm. The line AX intersects the line segment YZ at a right-angle at point A.

As XZ = XY = 5 cm, this implies that point A lies on the perpendicular bisector of YZ by the Converse of the Perpendicular Bisector Theorem. Thus, AY = AZ. Solving for x, we obtain,

**$2x-1=x+1\phantom{\rule{0ex}{0ex}}\Rightarrow 2x-x=1+1\phantom{\rule{0ex}{0ex}}\Rightarrow x=2$**

Now that we have found the value of x, we can calculate the side AY as

$\overline{)AY}=2x-1\phantom{\rule{0ex}{0ex}}\Rightarrow \overline{)AY}=2\left(2\right)-1\phantom{\rule{0ex}{0ex}}\Rightarrow \overline{)AY}=3$

Since AY = AZ , therefore, AY = AZ = 3 cm.

The **perpendicular bisector of a triangle** is a line segment that is drawn from the side of a triangle to the opposite vertex. This line is perpendicular to that side and passes through the midpoint of the triangle. The perpendicular bisector of a triangle divides the sides into two equal parts.

Every triangle has three perpendicular bisectors since it has three sides.

The **circumcenter **is a point at which all three perpendicular bisectors of a triangle intersect.

The circumcenter is the point of concurrency of the three perpendicular bisectors of a given triangle.

A point at which three or more distinct lines intersect is called a **point of concurrency**. Similarly, three or more lines are said to be concurrent if they pass through an identical point.

This is described in the diagram below where P is the circumcenter of the given triangle.

The vertices of a triangle are equidistant from the circumcenter. In other words, given a triangle ABC, if the perpendicular bisectors of AB, BC, and AC meet at point P, then AP = BP = CP.

**Proof**

Observe the triangle ABC above. The perpendicular bisectors of line segments AB, BC, and AC are given. The perpendicular bisector of AC and BC intersect at point P. We want to show that point P lies on the perpendicular bisector of AB and is equidistant from A, B, and C. Now observe the line segments AP, BP, and CP.

By the Perpendicular Bisector Theorem, any point on the perpendicular bisector is equidistant from both the endpoints of a line segment. Thus, AP = CP and CP = BP.

By the transitive property, AP = BP.

The transitive property states that if A = B and B = C, then A = C.

By the Converse of the Perpendicular Bisector Theorem, any point equidistant from the endpoints of a segment lies on the perpendicular bisector. Thus, P lies on the perpendicular bisector of AB. As AP = BP = CP, so point P is equidistant from A, B and C.

Say we are given three points, A, B, and C that make up a triangle on the Cartesian graph. To locate the circumcenter of the triangle ABC, we can follow the method below.

Evaluate the midpoint of the two sides.

Find the slope of the two chosen sides.

Calculate the slope of the perpendicular bisector of the two chosen sides.

Determine the equation of the perpendicular bisector of the two chosen sides.

Equate the two equations in Step 4 to each other to find the x-coordinate.

Plug the found x-coordinate into one of the equations in Step 4 to identify the y-coordinate.

Locate the coordinates of the circumcenter of the triangle XYZ given the vertices X (-1, 3), Y (0, 2), and Z (-2, -2).

Let us begin by sketching the triangle XYZ.

We shall attempt to find the perpendicular bisectors of the line segments XY and XZ given their respective midpoints.

**Perpendicular Bisector of XY**

The midpoint is given by:

$({x}_{XY},{y}_{XY})=\left(\frac{-1+0}{2},\frac{3+2}{2}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow ({x}_{XY},{y}_{XY})=\left(-\frac{1}{2},\frac{5}{2}\right)$

The slope of the line segment XY is:

${m}_{XY}=\frac{2-3}{0-(-1)}\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{XY}=-\frac{1}{1}\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{XY}=-1$

The slope of the perpendicular bisector of this line segment is:

$-1\times {m}_{pXY}=-1\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{pXY}=1$

We thus obtain the equation of the perpendicular bisector as

$y-{y}_{XY}={m}_{pXY}\left(x-{x}_{XY}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow y-\frac{5}{2}=x+\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow y=x+\frac{1}{2}+\frac{5}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow y=x+3$

**Perpendicular Bisector of**** **** XZ**

The midpoint is given by:

$({x}_{XZ},{y}_{XZ})=\left(\frac{-1+(-2)}{2},\frac{3+(-2)}{2}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow ({x}_{XZ},{y}_{XZ})=\left(-\frac{3}{2},\frac{1}{2}\right)$

The slope of the line segment XZ is:

${m}_{XZ}=\frac{-2-3}{-2-(-1)}\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{XZ}=\frac{-5}{-1}\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{XZ}=5$

The slope of the perpendicular bisector of this line segment is:

$5\times {m}_{pXZ}=-1\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{pXZ}=-\frac{1}{5}$

We thus obtain the equation of the perpendicular bisector as:

$y-{y}_{XZ}={m}_{pXZ}\left(x-{x}_{XZ}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow y-\frac{1}{2}=-\frac{1}{5}\left(x+\frac{3}{2}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow y=-\frac{1}{5}x-\frac{3}{10}+\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow y=-\frac{1}{5}x+\frac{1}{5}$

**Set the equations of the Perpendicular Bisector of XY = Perpendicular Bisector of XZ**

The x-coordinate is obtained by:

$x+3=-\frac{1}{5}x+\frac{1}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow x+\frac{1}{5}x=\frac{1}{5}-3\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{6}{5}x=-\frac{14}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow 6x=-14\phantom{\rule{0ex}{0ex}}\Rightarrow x=-\frac{7}{3}$

The y-coordinate can be found by:

$y=x+3\phantom{\rule{0ex}{0ex}}\Rightarrow y=-\frac{7}{3}+3\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{2}{3}$

Thus, the circumcenter is given by the coordinates$P\left(-\frac{7}{3},\frac{2}{3}\right)$

The Angle Bisector Theorem tells us that if a point lies on the bisector of an angle, then the point is equidistant from the sides of the angle.

This is described in the diagram below.

If the line segment CD bisects the ∠C and AD is perpendicular to AC and BD is perpendicular to BC, then AD = BD.

Before we begin the proof, recall the ASA Congruence rule.

**ASA Congruence**

If two angles and an included side of one triangle are equal to two angles and an included side of another triangle, then the triangles are congruent.

**Proof**

We need to show that AD = BD.

As the line CD bisects ∠C, this forms two angles of equal measures, namely ∠ACD = ∠BCD. Further, notice that since AD is perpendicular to AC and BD is perpendicular to BC, then ∠A = ∠B = 90^{o}. Finally, CD = CD for both triangles ACD and BCD.

By the ASA Congruence rule, Triangle ACD is congruent to Triangle BCD. Thus, AD = BD.

We can indeed use this theorem in the context of triangles. Applying this concept, the angle bisector of any angle in a triangle divides the opposite side into two parts that are proportional to the other two sides of the triangle. This angle bisector divides the bisected angle into two angles of equal measures.

This ratio is described in the diagram below for triangle ABC.

If the angle bisector of ∠C is represented by the line segment CD and ∠ACD = ∠BCD, then:

$\frac{\overline{)AC}}{\overline{)BC}}=\frac{\overline{)AD}}{\overline{)BD}}$

The Converse of the Angle Bisector Theorem states that if a point is equidistant from the sides of an angle, then the point lies on the bisector of the angle.

This is illustrated in the diagram below.

If AD is perpendicular to AC and BD is perpendicular to BC and AD = BD, then the line segment CD bisects the ∠C.

**Proof**

We need to show that CD bisects ∠C.

As AD is perpendicular to AC and BD is perpendicular to BC, then ∠A = ∠B = 90^{o}. We are also given that AD = BD. Lastly, both triangles ACD and BCD share a common side upon drawing a line segment through ∠C, that is, CD = CD.

By the SAS Congruence rule, Triangle ACD is congruent to Triangle BCD. Thus, CD bisects ∠C.

As before, we can apply this theorem to triangles as well. In this context, a line segment constructed from any angle of a triangle that divides the opposite side into two parts such that they are proportional to the other two sides of a triangle implies that the point on the opposite side of that angle lies on the angle bisector.

This concept is illustrated below for triangle ABC.

If $\frac{\overline{)AC}}{\overline{)BC}}=\frac{\overline{)AD}}{BD}$then D lies on the angle bisector of ∠C and the line segment CD is the angle bisector of ∠C.

Observe the triangle XYZ below.

Find the length of the side XZ if XA is the angle bisector of ∠X, XY = 8cm, AY = 3 cm and AZ = 4cm.

By the Angle Bisector Theorem for triangles, given that XA is the angle bisector of ∠X then

$\frac{\overline{)AY}}{\overline{)AZ}}=\frac{\overline{)XY}}{XZ}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3}{4}=\frac{8}{XZ}\phantom{\rule{0ex}{0ex}}\Rightarrow \overline{)XZ}=\frac{8\times 4}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \overline{)XZ}=\frac{32}{3}\approx 10.67(correcttotwodecimalplaces)$

Thus, the length of XZ is approximately 10.67 cm.

The same concept applies to the Converse of the Angle Bisector Theorem for triangles. Say we were given the triangle above with the measures XY = 8cm, XZ = $\frac{32}{3}$cm, AY = 3 cm and AZ = 4cm. We want to determine whether point A lies on the angle bisector of ∠X. Evaluating the ratio of the corresponding sides, we find that

$\frac{\overline{)AY}}{\overline{)AZ}}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{\overline{)XY}}{\overline{)XZ}}=\frac{8}{\frac{32}{3}}=\frac{3\times 8}{32}=\frac{24}{32}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\overline{)AY}}{\overline{)AZ}}=\frac{\overline{)XY}}{\overline{)XZ}}$

Thus, point A indeed lies on the angle bisector of ∠X and the line segment XA is the angle bisector of ∠X.

The** angle bisector of a triangle** is a line segment that is drawn from the vertex of a triangle to the opposite side. The angle bisector of a triangle divides the bisected angle into two equal measures.

Every triangle has three angle bisectors since it has three angles.

The **incenter **is a point at which all three angle bisectors of a triangle intersect.

The incenter is the point of concurrency of the three angle bisectors of a given triangle. This is illustrated in the diagram below where Q is the incenter of the given triangle.

** **

The sides of a triangle are equidistant from the incenter. In other words, given a triangle ABC, if the angle bisectors of ∠A, ∠B, and ∠C meet at point Q, then QX = QY = QZ.

**Proof**

Observe the triangle ABC above. The angle bisectors of ∠A, ∠B and ∠C are given. The angle bisector of ∠A and ∠B intersect at point Q. We want to show that point Q lies on the angle bisector of ∠C and is equidistant from X, Y and Z. Now observe the line segments AQ, BQ and CQ.

By the Angle Bisector Theorem, any point lying on the bisector of an angle is equidistant from the sides of the angle. Thus, QX = QZ and QY = QZ.

By the transitive property, QX = QY.

By the Converse of the Angle Bisector Theorem, a point that is equidistant from the sides of an angle lies on the bisector of the angle. Thus, Q lies on the angle bisector of ∠C. As QX = QY = QZ, so point Q is equidistant from X, Y and Z.

If s the incenter of the triangle XYZ, then find the value of in the figure below. XA, YB and ZC are the angle bisectors of the triangle.

∠YXA and ∠ZYB are given by 32^{o} and 27^{o} respectively. Recall that an angle bisector divides an angle into two equal measures. Further note that the sum of the interior angles of a triangle is 180^{o}.

Since Q is the incenter XA, YB and ZC are the angle bisectors of the triangle, then

$\angle YXA+\angle ZYB+\angle YZC=\frac{180}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {32}^{o}+{27}^{o}+\angle \theta ={90}^{o}\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \theta ={90}^{o}-{32}^{o}-{27}^{o}\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \theta ={90}^{o}-{32}^{o}-{27}^{o}\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \theta ={31}^{o}$

Thus, ∠θ = 31^{o}

The **median **is a line segment that connects the vertex of a triangle to the midpoint of the opposite side.

Every triangle has three medians since it has three vertices.

The **centroid **is a point at which all three medians of a triangle intersect.

The centroid is the point of concurrency of the three medians of a given triangle. This is shown in the illustration below where R is the incenter of the given triangle.

The centroid of a triangle is two-thirds of the distance from each vertex to the midpoint of the opposite side. In other words, given a triangle ABC, if the medians of AB, BC, and AC meet at a point R, then

$\overline{)AR}=\frac{2}{3}\overline{)AX}\phantom{\rule{0ex}{0ex}}\overline{)BR}=\frac{2}{3}\overline{)BY}\phantom{\rule{0ex}{0ex}}\overline{)CR}=\frac{2}{3}\overline{)CZ}$

If R is the centroid of the triangle XYZ, then find the value of AR and XR given that XA = 21 cm in the diagram below. XA, YB, and ZC are the medians of the triangle.

By the Centroid Theorem, we deduce that XR can be found by the formula:

$\overline{)XR}=\frac{2}{3}\overline{)XA}\phantom{\rule{0ex}{0ex}}\Rightarrow \overline{)XR}=\frac{2}{3}\left(21\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \overline{)XR}=14$

The value of AR is:

$\overline{)AR}+\overline{)XR}=\overline{)XA}\phantom{\rule{0ex}{0ex}}\Rightarrow \overline{)AR}\hspace{0.17em}+\hspace{0.17em}14=21\phantom{\rule{0ex}{0ex}}\Rightarrow \overline{)AR}=21-15\phantom{\rule{0ex}{0ex}}\Rightarrow \overline{)AR}=7$

Thus, $\overline{)XR}=14$ cm and $\overline{)AR}=7$cm.

The **altitude **is a line segment that passes through the vertex of a triangle and is perpendicular to the opposite side.

Every triangle has three altitudes since it has three vertices.

The **orthocenter **is a point at which all three altitudes of a triangle intersect.

The orthocenter is the point of concurrency of the three altitudes of a given triangle. This is described in the image below where S is the orthocenter of the given triangle.

It may be helpful to note that the location of the orthocenter, S depends on the type of triangle given.

Type of Triangle | Position of the Orthocenter, S |

Acute | S lies inside the triangle |

Right | S lies on the triangle |

Obtuse | S lies outside the triangle |

Say we are given a set of three points for a given triangle A, B and C. We can determine the coordinates of the orthocenter of a triangle using the Orthocenter Formula. This is given by the technique below.

Find the slope of the two sides

Calculate the slope of the perpendicular bisector of the two chosen sides (note that the altitude for each vertex of the triangle coincides with the opposite side).

Determine the equation of the perpendicular bisector of the two chosen sides with its corresponding vertex.

Equate the two equations in Step 3 to each other to find the x-coordinate.

Plug the found x-coordinate into one of the equations in Step 3 to identify the y-coordinate.

Locate the coordinates of the orthocenter of the triangle XYZ given the vertices X (-5, 7), Y (5, -1), and Z (-3, 1). XA, YB and ZC are the altitudes of the triangle.

We begin by drawing a rough sketch of the triangle XYZ.

Example 7, Aishah Amri - StudySmarter Originals

We shall attempt to find the perpendicular bisectors of the line segments XY and XZ given their respective vertices.

**Perpendicular Bisector of XY**

The corresponding vertex for XY is given by the point Z (-3, 1)

The slope of the line segment XY is:

${m}_{XY}=\frac{-1-7}{5-(-5)}\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{XY}=-\frac{8}{10}\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{XY}=-\frac{4}{5}$

The slope of the perpendicular bisector of this line segment is:

$-\frac{4}{5}\times {m}_{pXY}=-1\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{pXY}=\frac{5}{4}$

We thus obtain the equation of the perpendicular bisector as:

$y-{y}_{Z}={m}_{pXY}\left(x-{x}_{Z}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow y-1=\frac{5}{4}\left(x+3\right)\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{5}{4}x+\frac{15}{4}+1\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{5}{4}x+\frac{19}{4}$

**Perpendicular Bisector of**** **** XZ**

The corresponding vertex for XZ is given by the point Y (5, -1)

The slope of the line segment XZ is:

${m}_{XZ}=\frac{1-7}{-3-(-5)}\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{XZ}=-\frac{6}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{XZ}=-3$

The slope of the perpendicular bisector of this line segment is:

$-3\times {m}_{pXZ}=-1\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{pXZ}=\frac{1}{3}$

We thus obtain the equation of the perpendicular bisector as:

$y-{y}_{Y}={m}_{pXZ}\left(x-{x}_{Y}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow y+1=\frac{1}{3}\left(x-5\right)\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{1}{3}x-\frac{5}{3}-1\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{1}{3}x-\frac{8}{3}$

**Set the equations of the Perpendicular Bisector of XY = Perpendicular Bisector of XZ**

The x-coordinate is obtained by:

$\frac{5}{4}x+\frac{19}{4}=\frac{1}{3}x-\frac{8}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{5}{4}x-\frac{1}{3}x=-\frac{8}{3}-\frac{19}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{11}{12}x=-\frac{89}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow 11x=-89\phantom{\rule{0ex}{0ex}}\Rightarrow x=-\frac{89}{11}$

The y-coordinate can be found by:

$y=\frac{5}{4}x+\frac{19}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{5}{4}\left(-\frac{89}{11}\right)+\frac{19}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow y=-\frac{59}{11}$

Thus, the orthocenter is given by the coordinates $S\left(-\frac{89}{11},-\frac{59}{11}\right)$

**Important Theorems**TheoremDescriptionThe Perpendicular Bisector TheoremAny point on the perpendicular bisector is equidistant from both the endpoints of a line segment.

The Converse of the Perpendicular Bisector TheoremIf a point is equidistant from the endpoints of a line segment in the same plane, then that point lies on the perpendicular bisector of the line segment.

The Angle Bisector TheoremIf a point lies on the bisector of an angle, then the point is equidistant from the sides of the angle.

The Angle Bisector Theorem and TrianglesThe angle bisector of any angle in a triangle divides the opposite side into two parts that are proportional to the other two sides of the triangle and divides the bisected angle into two angles of equal measures.

The Converse of the Angle Bisector TheoremIf a point is equidistant from the sides of an angle, then the point lies on the bisector of the angle.

The Converse of the Angle Bisector Theorem and TrianglesA line segment constructed from any angle of a triangle that divides the opposite side into two parts such that they are proportional to the other two sides of a triangle implies that the point on the opposite side of that angle lies on the angle bisector.**Important Concepts**ConceptPoint of ConcurrencyPropertyPerpendicular bisectorCircumcenterThe vertices of a triangle are equidistant from the circumcenter.Angle bisectorIncenterThe sides of a triangle are equidistant from the incenter.CentroidThe centroid of a triangle is two-thirds of the distance from each vertex to the midpoint of the opposite side.OrthocenterThe line segments including the altitudes of the triangle are concurrent at the orthocenter.**Method**: Determine the Equation of the Perpendicular Bisector- Find the coordinates of the midpoint.
- Calculate the slope of the chosen line segments.
- Determine the slope of the perpendicular bisector.
- Evaluate the equation of the perpendicular bisector.

**Method**: Finding the Coordinates of the Circumcenter of a TriangleEvaluate the midpoint of two sides.

Find the slope of the two chosen sides.

Calculate the slope of the perpendicular bisector of the two chosen sides.

Determine the equation of the perpendicular bisector of the two chosen sides.

Equate the two equations in Step 4 to each other to find the x-coordinate.

Plug the found x-coordinate into one of the equations in Step 4 to identify the y-coordinate.

**Method**: Locating the Orthocenter of a Triangle- Find the slope of the two sides.
- Calculate the slope of the perpendicular bisector of the two chosen sides.
- Determine the equation of the perpendicular bisector of the two chosen sides with its corresponding vertex.
- Equate the two equations in Step 3 to each other to find the x-coordinate.
- Plug the found x-coordinate into one of the equations in Step 3 to identify the y-coordinate.

The perpendicular bisector divides a segment into two equal halves.

How to find the equation of a perpendicular bisector:

- Find the midpoint of two given points
- Calculate the slope of two given points
- Derive the slope of the perpendicular bisector
- Determine the equation of the perpendicular bisector

^{o}. The perpendicular bisector divides the intersected line into two equal parts at its midpoint.

More about Perpendicular Bisector

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