Analytic Geometry · Analytic Geometry

Equations of Right Bisectors, Medians and Altitudes

2.5About 75 minutes

Learning goals

I can find the equation of the right bisector of a line segment.
I can find equations of medians and altitudes of a triangle.
I can use the right bisector to find points that are equidistant from two points.
I can explain the difference between a median, an altitude and a right bisector.

Words to know

right bisector
The line through the midpoint of a segment that is perpendicular to it (also called the perpendicular bisector).

median
A segment from a vertex of a triangle to the midpoint of the opposite side.

altitude
A segment from a vertex of a triangle perpendicular to the line containing the opposite side.

negative reciprocal
For a/b, the number −b/a. Perpendicular lines have slopes that are negative reciprocals.

equidistant
The same distance from two or more points.

circumcentre
The point where the right bisectors of a triangle's sides meet; it is equidistant from the three vertices.

Equal distances

Two towns, A and B, want to build a shared recycling depot on a road so that it is exactly the same distance from each town. Where could it go? Every point that is equidistant from A and B lies on one special line: the right bisector of segment AB. In this lesson you will find equations of right bisectors and other important lines in triangles by combining slope, midpoint and length.

Remember from Grade 9

Slope m = rise ÷ run = (y₂ − y₁)/(x₂ − x₁).
Parallel lines have equal slopes.
Perpendicular lines have slopes that are negative reciprocals: if one slope is a/b, the other is −b/a. Their product is −1 (for lines that are not vertical or horizontal).
To find the equation of a line through a point with a known slope, substitute the point into y = mx + b and solve for b.

Three special lines

The right bisector (also called the perpendicular bisector) of a segment passes through the segment's midpoint and is perpendicular to it. Every point on it is the same distance from both endpoints.
A median of a triangle joins a vertex to the midpoint of the opposite side.
An altitude of a triangle passes through a vertex and is perpendicular to the opposite side (it measures the triangle's height from that side).

To find any of these equations you need two things: a point on the line and its slope.

Line	Point to use	Slope to use
Right bisector of AB	Midpoint of AB	Negative reciprocal of the slope of AB
Median from vertex C	C and the midpoint of AB	Slope between those two points
Altitude from vertex C	C	Negative reciprocal of the slope of AB

What you need for each line.

Worked example

Example 1: Equation of a right bisector

Find the equation of the right bisector of the segment with endpoints A(−2, 1) and B(4, 5).

Midpoint: M = ((−2 + 4)/2, (1 + 5)/2) = (1, 3).
Slope of AB: (5 − 1)/(4 − (−2)) = 4/6 = 2/3.
Perpendicular slope: the negative reciprocal of 2/3 is −3/2.
Equation: y = −(3/2)x + b through (1, 3): 3 = −3/2 + b, so b = 3 + 3/2 = 9/2.
So y = −(3/2)x + 9/2. In standard form (multiply by 2 and rearrange): 3x + 2y − 9 = 0.
Check equidistance with a point on the line, e.g. x = −1: y = 3/2 + 9/2 = 6, so P(−1, 6). PA = √(1 + 25) = √26 and PB = √(25 + 1) = √26 ✓.

The right bisector is y = −1.5x + 4.5 (or 3x + 2y − 9 = 0).

The dashed right bisector passes through the midpoint M at a right angle. The double tick marks show AM = MB. Any point on it, such as P(−1, 6), is the same distance from A and B (the two thin segments are equal, √26 each).

Worked example

Example 2: Equation of a median

Triangle PQR has vertices P(−3, −2), Q(5, 6) and R(7, −4). Find the equation of the median from R.

The median from R goes to the midpoint of the opposite side, PQ: N = ((−3 + 5)/2, (−2 + 6)/2) = (1, 2).
Slope RN = (2 − (−4))/(1 − 7) = 6/(−6) = −1.
y = −x + b through (1, 2): 2 = −1 + b, b = 3.
Check with R: −7 + 3 = −4 ✓.

The median from R is y = −x + 3.

Worked example

Example 3: Equation of an altitude

For the same triangle, find the equation of the altitude from P.

The altitude from P is perpendicular to the opposite side, QR.
Slope QR = (−4 − 6)/(7 − 5) = −10/2 = −5.
Perpendicular slope: the negative reciprocal of −5 is 1/5.
y = (1/5)x + b through P(−3, −2): −2 = −3/5 + b, b = −2 + 3/5 = −7/5.
Check: (1/5)(−3) − 7/5 = −3/5 − 7/5 = −10/5 = −2 ✓.

The altitude from P is y = (1/5)x − 7/5, or x − 5y − 7 = 0.

In triangle PQR, the dashed median joins R to N, the midpoint of PQ (tick marks show PN = NQ). The solid altitude from P meets QR at a right angle. Notice that the altitude does not go to the midpoint of QR.

Do not mix them up

A median goes to a midpoint (but usually not at a right angle). An altitude makes a right angle with the opposite side (but usually not at its midpoint). A right bisector does both, but it does not have to pass through a vertex. In an equilateral triangle, all three are the same line.

Worked example

Example 4: A fair location

On a map grid (units in km), the villages of Elmvale at E(1, 7) and Fairhaven at F(9, 3) want a new arena on Highway 12, which runs along the line y = x − 4. The arena must be the same distance from both villages. Where should it go?

Points equidistant from E and F are on the right bisector of EF.
Midpoint of EF: (5, 5). Slope EF = (3 − 7)/(9 − 1) = −4/8 = −1/2. Perpendicular slope: 2.
Right bisector: y = 2x + b through (5, 5): b = −5. So y = 2x − 5.
Intersect with the highway (a linear system!): 2x − 5 = x − 4, so x = 1, y = −3.
Check distances from (1, −3): to E: √(0 + 100) = 10 km. To F: √(64 + 36) = √100 = 10 km ✓.

Build the arena at (1, −3), 10 km from each village. (The map coordinates are made up for this example and do not match the real places.)

Just for fun

Why are perpendicular lines such good friends?

Because every time they meet, they're right.

Hands-on activity

Fold the three special lines

You need: a sheet of paper, a ruler, a pencil, and scissors (grown-up note: take care with scissors).

Draw a large triangle that is clearly not equilateral and not right-angled. Cut it out. Label the vertices A, B and C.
Right bisector of AB: fold so that A lands exactly on B and crease. Unfold and draw the crease in pencil.
Median from C: pinch the middle of AB (where the crease crossed it). Fold through C and that midpoint. Draw it with a different style (dashed).
Altitude from C: fold side AB onto itself so the crease passes through C. (The edge AB lines up with itself; the crease is then perpendicular to AB.) Draw it with a third style (dotted).
Which of your three lines pass through C? Which pass through the midpoint of AB? Which make a right angle with AB?

What to look for: the right bisector goes through the midpoint at 90° but usually misses C; the median goes through C and the midpoint; the altitude goes through C at 90° but usually misses the midpoint. If your triangle were isosceles with CA = CB, all three would be the same crease.

Your turn

Find the slope of a line perpendicular to a line with slope 3/4.

Find the equation of the right bisector of the segment from (0, 0) to (6, 4).

Find the equation of the right bisector of the segment from (−4, 3) to (2, 3).

Triangle ABC has A(0, 6), B(−4, 0) and C(6, 2). Find the equation of the median from A.

For the same triangle, find the equation of the altitude from A.

Find the equation of the line through (2, −1) parallel to y = 4x + 7.

More practice: warm-up

What is the negative reciprocal of −2?

What is the slope of the segment from (1, 2) to (5, 10)?

A median joins a vertex to the ___ of the opposite side.

An altitude is ___ to the opposite side.

Find b if y = 3x + b passes through (2, 11).

More practice: on your own

Find the right bisector of the segment from (−1, −5) to (5, 3). Write it in standard form.

Triangle DEF has D(−2, 4), E(4, 6) and F(2, −2). Find the equation of the altitude from F.

For the same triangle, find the equation of the median from D.

Is the point (5, 1) on the right bisector of the segment from (2, 3) to (6, 7)? Answer two ways: with distances, and with the equation.

Find the point on the y-axis that is equidistant from (2, 2) and (6, −2).

Challenge

Triangle ABC has A(−4, 2), B(4, 6) and C(2, −4). Find the equations of two of its medians and their intersection (the centroid). Compare the centroid with the average of the three vertices.

Find the circumcentre of the triangle with vertices O(0, 0), A(8, 0) and B(2, 6) by intersecting two right bisectors. Check that it is equidistant from all three vertices.

Explain why every point on the right bisector of AB is equidistant from A and B. (Hint: think about two triangles.)

Check your understanding

Quiz

The slope of AB is −2/5. What is the slope of the right bisector of AB?

A median of a triangle joins a vertex to:

What is the midpoint you need for the right bisector of (−3, 4) and (5, −2)?

Which line is the altitude from (0, 0) to the side joining (4, 0) and (4, 6)?

Every point on the right bisector of AB is:

Which equation is the right bisector of the segment from (0, 0) to (4, 4)?

For parents and teachers

Big idea: to write the equation of a line you need a point and a slope; right bisectors, medians and altitudes each get these from midpoints and (perpendicular) slopes. This lesson teaches expectation 2.5 (solve problems involving the slope, length and midpoint of a line segment, such as finding the equation of the right bisector).

How to help: Ask, "Which point? Which slope?" for each line. The folding activity shows the difference between the three special lines very clearly.

Common difficulties:

Using the slope of the segment instead of the perpendicular slope for a right bisector or altitude.
Taking only the reciprocal or only the negative (the perpendicular slope to 2/3 is −3/2).
Confusing medians and altitudes.
Horizontal and vertical segments: the right bisector of a horizontal segment is a vertical line x = a, which has no slope.