VertexMath · MPM2D · Free worksheet

Analytic Geometry Review

Try each problem first. You don’t need to finish everything.

Core path ~55 min Full sheet ~1 h 45 min

Recommended: K1–K5, then one A and one T problem

K = Knowledge & Understanding · A = Application · T = Thinking · Boss = Challenge

vertexmath.ca Hints, answers & more worksheets

Quick reference

Midpoint: $M=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$ . Length: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ , and the square root is part of the answer.
Circle centred at the origin with radius $r$ : $x^2+y^2=r^2$ . The number on the right is $r^2$ , not $r$ .
Slopes: parallel lines have equal slopes; perpendicular lines have slopes whose product is $-1$ . The shortest distance from a point to a line is measured along the perpendicular.
Median: vertex to the midpoint of the opposite side. Altitude: from a vertex, perpendicular to the opposite side. Perpendicular bisector: through a side's midpoint, perpendicular to it. Three different lines.
Four equal sides make a rhombus. It is a square only if adjacent sides are also perpendicular.

Show your work on separate paper. Show your work in the dot-grid space.

Core Skills (K)

Midpoint & Length · Circles · Classifying Figures · Special Lines · Point-to-Line Distance

K1 Midpoint & Length

a)
Determine the midpoint of the segment joining $A(-3,4)$ and $B(5,-2)$ .
b)
Determine the length of the segment joining $C(-1,-3)$ and $D(3,5)$ , in simplest radical form.
c)
$M(2,-1)$ is the midpoint of $PQ$ , and $P$ is $(-4,3)$ . Find the coordinates of $Q$ .

Stuck? Peek at a hint

The midpoint averages the coordinates; the length squares the differences before taking the root. In (c), the step from $P$ to $M$ repeats from $M$ to $Q$ .

K2 Circles at the Origin

a)
State the radius and the four intercepts of the circle $x^2+y^2=45$ .
b)
A circle centred at the origin has a diameter with endpoints $(-5,12)$ and $(5,-12)$ . Determine its equation.
c)
Determine whether each point lies inside, on, or outside the circle $x^2+y^2=45$ : $(6,-3)$ , $(-4,5)$ , $(5,5)$ .

Stuck? Peek at a hint

The number on the right side of $x^2+y^2=r^2$ is the radius squared. For (c), substitute each point into $x^2+y^2$ and compare with $45$ .

K3 Classify the Triangle

$\triangle ABC$ has vertices $A(-2,1)$ , $B(2,3)$ and $C(4,-1)$ . Each grid square is 1 unit.

a)
Determine the exact length of each side.
b)
Determine the slope of each side.
c)
Classify $\triangle ABC$ by its sides and by its angles.

Stuck? Peek at a hint

Use the distance formula for (a) and rise over run for (b). Two sides are perpendicular when their slopes multiply to $-1$ .

K4 Three Special Lines

$\triangle ABC$ has vertices $A(-4,0)$ , $B(2,6)$ and $C(4,4)$ . Write each equation in the form $y=mx+b$ .

a)
Determine the equation of the median from $B$ .
b)
Determine the equation of the altitude from $B$ .
c)
Determine the equation of the perpendicular bisector of $AC$ .

Stuck? Peek at a hint

Find the midpoint and the slope of $AC$ first. Each line uses one of them, or both.

K5 Point to Line

Line $\ell$ has equation $y=2x-3$ , and $P$ is the point $(6,-1)$ .

a)
Determine the equation of the line through $P$ perpendicular to $\ell$ .
b)
Find the point where this line meets $\ell$ .
c)
Determine the shortest distance from $P$ to $\ell$ , exactly and to one decimal place.

Stuck? Peek at a hint

The shortest path from a point to a line meets the line at a right angle.

Real World (A)

Circles in Context · Shortest Distance · Equidistant Points

A1 Cell Tower Coverage

A cell tower in rural Northern Ontario stands at the origin of a map where each unit is 1 km. Its signal reaches every point within 13 km of the tower.

a)
Write the equation of the boundary of the coverage area.
b)
Cottages stand at $(5,-12)$ and $(-9,10)$ . Determine whether each cottage has coverage.
c)
A new road starts at $(0,5)$ and runs due east along the line $y=5$ . How far along the road can a driver travel before losing coverage?

Stuck? Peek at a hint

Every point on the boundary is exactly 13 km from the origin. For (c), substitute $y=5$ into your equation.

A2 Shortest Driveway

On a farm map, each grid square is 1 unit and each unit is 10 m. A straight road runs from $R(0,2)$ to a gate at $E(8,6)$ , where it ends. A house stands at $H(1,10)$ .

a)
Find the equation of the road's line, then the point on the road closest to $H$ .
b)
Determine the length of the shortest driveway from $H$ to the road, in metres, to the nearest tenth.
c)
A second house stands at $G(12,3)$ . Determine the length of the shortest driveway from $G$ to the road.

Stuck? Peek at a hint

The shortest driveway meets the road at a right angle, so its slope is the negative reciprocal of the road's slope.

A3 One Lamp for the Park

The corners of a triangular park are given in the table, where each unit is 10 m. The city wants one lamp post that is the same distance from all three corners.

$\text{Corner}$	$A$	$B$	$C$
$x$	$-6$	$6$	$2$
$y$	$0$	$0$	$4$

a)
Determine the coordinates of the point that is the same distance from $A$ , $B$ and $C$ .
b)
Determine that distance in metres, to the nearest tenth.
c)
The lamp must stand inside the park. Can the city put it at your point from (a)? What does this mean for the plan?

Stuck? Peek at a hint

A point the same distance from $A$ and $B$ lies on the perpendicular bisector of $AB$ . Intersect two perpendicular bisectors.

Think It Through (T)

Error Analysis · Explain Why

T1 Square or Not?

Quadrilateral $PQRS$ has vertices $P(-3,0)$ , $Q(1,3)$ , $R(5,0)$ and $S(1,-3)$ . A student writes:
" $PQ=QR=RS=SP=5$ , so $PQRS$ is a square."

a)
Determine the slopes of $PQ$ and $QR$ . Use them to test the student's conclusion.
b)
Classify $PQRS$ as precisely as possible, and verify one property of its diagonals.
c)
Keep $P$ and $R$ fixed. Find new coordinates for $Q$ and $S$ that make $PQRS$ a square.

Stuck? Peek at a hint

A square needs four equal sides and a right angle. Check the second condition with slopes.

T2 Prove It for Every Triangle

$\triangle OAB$ has vertices $O(0,0)$ , $A(2a,0)$ and $B(2b,2c)$ , with $a>0$ and $c>0$ . $M$ is the midpoint of $OB$ and $N$ is the midpoint of $AB$ .

a)
Determine the coordinates of $M$ and $N$ .
b)
Show that $MN$ is parallel to $OA$ and half its length.
c)
A student checks the triangle with vertices $(0,0)$ , $(4,0)$ and $(2,6)$ and says this proves the property for all triangles. Explain why that check is not a proof, and why the coordinates in (b) cover every triangle.

Stuck? Peek at a hint

Apply the midpoint formula to the letters exactly as you would to numbers.

Boss Level

Optional challenge

Lattice Points on a Circle

A lattice point is a point whose coordinates are both integers.

a)
List all lattice points on the circle $x^2+y^2=25$ .
b)
How many lattice points lie on the circle $x^2+y^2=65$ ?
c)
How many lattice points lie strictly inside the circle $x^2+y^2=10$ ?

Stuck? Peek at a hint

Write the right side as a sum of two perfect squares in every possible way, then attach signs and swap the order. For (c), work column by column: for each integer $x$ , count the integers $y$ with $x^2+y^2<10$ .

Answer key

Final answers only. Try first, then check.

Core Skills

a) $(1,1)$
b) $\sqrt{80}=4\sqrt{5}$
c) $Q(8,-5)$

a) $r=3\sqrt{5}\approx 6.7$ ; intercepts $(3\sqrt{5},0)$ , $(-3\sqrt{5},0)$ , $(0,3\sqrt{5})$ , $(0,-3\sqrt{5})$
b) $x^2+y^2=169$
c) $(6,-3)$ : on ( $45$ ); $(-4,5)$ : inside ( $41$ ); $(5,5)$ : outside ( $50$ )

a) $AB=2\sqrt{5}$ , $BC=2\sqrt{5}$ , $AC=2\sqrt{10}$
b) $m_{AB}=\frac{1}{2}$ , $m_{BC}=-2$ , $m_{AC}=-\frac{1}{3}$
c) Isosceles right triangle: $AB=BC$ , and $m_{AB}\cdot m_{BC}=-1$ , so the right angle is at $B$ .

a) $y=2x+2$
b) $y=-2x+10$
c) $y=-2x+2$

a) $y=-\frac{1}{2}x+2$
b) $(2,1)$
c) $2\sqrt{5}\approx 4.5$ units

Real World

a) $x^2+y^2=169$
b) $(5,-12)$ : exactly 13 km away, on the edge of coverage. $(-9,10)$ : $\sqrt{181}\approx 13.5$ km away, no coverage.
c) $12$ km. $x^2+25=169$ gives $x=\pm12$ , but the road only has $x\ge 0$ , so $x=-12$ is not on it.

a) $y=\frac{1}{2}x+2$ ; $(4,4)$
b) $3\sqrt{5}$ units $\approx 67.1$ m
c) $50$ m. The perpendicular from $G$ meets the road's line at $(10,7)$ , past the gate, so the shortest driveway goes to $E$ : $GE=5$ units. (The perpendicular distance, $\approx 44.7$ m, cannot be reached.)

a) $(0,-2)$
b) $2\sqrt{10}$ units $\approx 63.2$ m
c) No: $(0,-2)$ is 20 m south of side $AB$ , outside the park, because the angle at $C$ is obtuse. That point is the only one equidistant from all three corners, so no spot inside the park works; the plan needs to change.

Think It Through

a) $m_{PQ}=\frac{3}{4}$ , $m_{QR}=-\frac{3}{4}$ ; product $-\frac{9}{16}\ne-1$ , so the angle at $Q$ is not right: not a square.
b) Rhombus. Diagonals $PR$ (on $y=0$ ) and $QS$ (on $x=1$ ) are perpendicular and bisect each other at $(1,0)$ ; they have different lengths, $8$ and $6$ .
c) $Q(1,4)$ and $S(1,-4)$

a) $M(b,c)$ , $N(a+b,c)$
b) $m_{MN}=\frac{c-c}{a}=0=m_{OA}$ , so $MN\parallel OA$ ; $MN=a$ and $OA=2a$ .
c) One example shows the property for that triangle only. Any triangle can be placed with one vertex at the origin and one side along the positive $x$ -axis, so $O(0,0)$ , $A(2a,0)$ , $B(2b,2c)$ with any $a>0$ , $c>0$ and any $b$ describes every triangle.

Boss Level

Boss

a) 12 points: $(\pm5,0)$ , $(0,\pm5)$ , $(\pm3,\pm4)$ , $(\pm4,\pm3)$
b) $16$ : $(\pm1,\pm8)$ , $(\pm8,\pm1)$ , $(\pm4,\pm7)$ , $(\pm7,\pm4)$
c) $29$

Free cheat sheet

Analytic Geometry Cheat Sheet

The whole unit on one page: every rule, a worked example for each idea, and the traps that cost marks.

Unit Summary

One printable page with every rule.

↓ Unit Summary PDF

Build your own

The same page with blanks: fill it in from memory, then check.

↓ Build-your-own PDF

Phone card

The essentials on one screen. Save it to your photos.

↓ Phone card (image)

Most testedFind the shortest distance from $P(7,0)$ to the line $y=2x+1$ . (card 7)

1Midpoint and length

$M=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Example $A(-1,2)$ , $B(5,10)$ : $M(2,6)$ , $AB=\sqrt{36+64}=10$ .

TrapThe midpoint averages: $(-1+5)/2=2$ , not $(5-(-1))/2=3$ .

TrapMidpoint $M(2,6)$ and endpoint $A(-1,2)$ : $B=(2\cdot2+1,\ 2\cdot6-2)=(5,10)$ . Averaging $A$ and $M$ gives a quarter point.

2Circle at the origin

$x^2+y^2=r^2$

UseA point on it gives $r^2$ . Intercepts $(\pm r,0)$ , $(0,\pm r)$ . Inside: $x^2+y^2<r^2$ ; outside: $>r^2$ .

ExampleThrough $(8,-6)$ : $r^2=64+36=100$ , so $x^2+y^2=100$ and $r=10$ .

TipThe perpendicular bisector of any chord passes through the centre.

Trap $x^2+y^2=20$ has $r=2\sqrt5$ , not $20$ .

TrapDiameter from $(-6,8)$ to $(6,-8)$ : $r=10$ , so $x^2+y^2=100$ , not $20^2=400$ .

3Slopes and triangles

$m=\dfrac{y_2-y_1}{x_2-x_1}$

parallel: $m_1=m_2$

perpendicular: $m_1m_2=-1$

UseSides by lengths: scalene, isosceles, equilateral. Angles by slopes: a right angle needs $m_1m_2=-1$ .

Example $A(0,0)$ , $B(4,2)$ , $C(3,4)$ : $m_{AB}=1/2$ , $m_{BC}=-2$ , product $-1$ : right angle at $B$ .

Trap $AB=BC$ makes it isosceles, not right-angled. Test the slopes; never trust the picture.

4Quadrilaterals

one pair of parallel sides	trapezoid
both pairs parallel	parallelogram
parallelogram, a right angle	rectangle
four equal sides	rhombus
rhombus, a right angle	square

TrapFour equal sides make a rhombus. It is a square only if adjacent slopes multiply to $-1$ .

5Lines in a triangle

median: vertex to the midpoint of the opposite side	meet at the centroid
altitude: from a vertex, $\perp$ to the opposite side	orthocentre
perpendicular bisector: through a side's midpoint, $\perp$ to it	circumcentre

TrapOpposite side slope $3/4$ : the altitude's slope is $-4/3$ , not $3/4$ .

6Proving properties

UseMidsegment: parallel to the third side and half its length. The circumcentre is the same distance from all three vertices. A line from the centre $\perp$ to a chord bisects it.

TrapOne numerical triangle only verifies. A proof uses general vertices such as $(0,0)$ , $(2a,0)$ , $(2b,2c)$ .

TrapThe point equidistant from the vertices is the circumcentre, not the centroid. It can lie outside.

7Distance to a line

UsePerpendicular through $P$ (negative reciprocal slope); find where it meets the line; measure from $P$ to that point.

Example $P(7,0)$ , $y=2x+1$ : perpendicular $y=-x/2+3.5$ meets it at $(1,3)$ ; distance $\sqrt{36+9}=3\sqrt5\approx6.7$ .

TrapNot vertically: from $(7,0)$ straight up to $y=2x+1$ is $15$ , but the shortest distance is $3\sqrt5$ .

TrapOn a segment, check the foot lies on it; if not, use the nearest endpoint.

Which tool?

Is there a right angle? Product of slopes

Equal sides, perimeter, on a circle? Length formula

A point halfway, or a bisector Midpoint formula

↓ Worksheet PDF ↓ Answer key PDF

Google Classroom Email

More free MPM2D worksheets

Other courses

All MPM2D resources → · All free resources →

What is VertexMath?

VertexMath is daily math practice for Ontario high school students, with a tutor that asks questions instead of giving answers. These worksheets are free, whether or not you use it.

Visit vertexmath.ca →