Unit 2 · Ontario MPM2D

Ontario Grade 10 Analytic Geometry Practice Questions

Algebra on a grid. Midpoint and distance are the basics — medians, bisectors and altitudes build on them. This page covers midpoints, distances, and lines through triangles. Sketch quickly as you go.

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What you need to know

Midpoint — average the x's and the y's: ((x1+x2)/2, (y1+y2)/2).
Distance — √((x2−x1)²+(y2−y1)²), keep it exact until the end.
Median — midpoint of the opposite side, then line through the vertex.
Perpendicular bisector — midpoint plus negative reciprocal slope.

Practice questions with solutions

13 questions. Try each one on paper, then open the worked solution.

Question 1 · Medians & Perpendicular Bisectors

Find the equation of the median from vertex A for the triangle with vertices A(4,4), B(-6,2), and C(2,0). Write in slope y-intercept form.

Show worked solution

Median from A goes to midpoint of BC.
Midpoint of BC = ((-6+2)/2, (2+0)/2) = (-2, 1)

Slope from A(4,4) to midpoint(-2,1):
m = (1-4)/(-2-4) = -3/-6 = 1/2

y - 4 = 1/2(x - 4)
y = 1/2x - 2 + 4
y = 1/2x + 2

Answer: y = 1/2x + 2

Question 2 · Medians & Perpendicular Bisectors

Find the equation of the perpendicular bisector of line segment AB with endpoints A(-5,-2) and B(3,6). Write in slope y-intercept form.

Show worked solution

Midpoint of AB = ((-5+3)/2, (-2+6)/2) = (-1, 2)

Slope of AB = (6-(-2))/(3-(-5)) = 8/8 = 1

Slope of perpendicular bisector = -1 (negative reciprocal)

y - 2 = -1(x - (-1))
y - 2 = -(x + 1)
y = -x - 1 + 2
y = -x + 1

Answer: y = -x + 1

Question 3 · Medians & Perpendicular Bisectors

Find the equation of the perpendicular bisector of line segment PQ with endpoints P(8,8) and Q(4,-2). Write in slope y-intercept form.

Show worked solution

Midpoint of PQ = ((8+4)/2, (8+(-2))/2) = (6, 3)

Slope of PQ = (-2-8)/(4-8) = -10/-4 = 5/2

Slope of perpendicular bisector = -2/5 (negative reciprocal)

y - 3 = -2/5(x - 6)
y = -2/5x + 12/5 + 3
y = -2/5x + 12/5 + 15/5
y = -2/5x + 27/5

Answer: y = -2/5x + 27/5

Question 4 · Point-to-Line Shortest Distance Applications

A coastline is modeled by the line x+y=10x + y = 10x+y=10. A boat is at B(1,2)B(1, 2)B(1,2). Find the shortest distance from the boat to the coastline as an exact simplified radical.

Show worked solution

Line: x+y−10=0x + y - 10 = 0x+y−10=0. Slope of coastline is -1, so perpendicular slope is 1.
Perpendicular through BBB: y−2=1(x−1)⇒y=x+1y - 2 = 1(x - 1) \Rightarrow y = x + 1y−2=1(x−1)⇒y=x+1.
Intersect with x+y=10x + y = 10x+y=10: x+(x+1)=10⇒2x=9⇒x=4.5x + (x + 1) = 10 \Rightarrow 2x = 9 \Rightarrow x = 4.5x+(x+1)=10⇒2x=9⇒x=4.5, y=5.5y = 5.5y=5.5.d=(4.5−1)2+(5.5−2)2=3.52+3.52=3.52=722d = \sqrt{(4.5-1)^2 + (5.5-2)^2} = \sqrt{3.5^2 + 3.5^2} = 3.5\sqrt{2} = \frac{7}{2}\sqrt{2}d=(4.5−1)2+(5.5−2)2​=3.52+3.52​=3.52​=27​2​.
Therefore, the shortest distance is 722\dfrac{7\sqrt{2}}{2}272​​.

Answer: 7√2/2

Question 5 · Triangle Classification & Circles

Classify the triangle with vertices D(-5,2), E(2,5), and F(2,-1) as scalene, isosceles, or equilateral.

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DE = √((2-(-5))² + (5-2)²) = √(49+9) = √58
DF = √((2-(-5))² + (-1-2)²) = √(49+9) = √58
EF = √((2-2)² + (-1-5)²) = √(0+36) = 6

Since DE = DF = √58, the triangle is isosceles.

Answer: isosceles

Question 6 · Triangle Classification & Circles

Determine if triangle X(1,4) Y(-3,-2) Z(3,-6) is a right triangle.

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XY = √((-3-1)² + (-2-4)²) = √(16+36) = √52
XZ = √((3-1)² + (-6-4)²) = √(4+100) = √104
YZ = √((3-(-3))² + (-6-(-2))²) = √(36+16) = √52

Check: XY² + YZ² = 52 + 52 = 104 = XZ²
Since XY² + YZ² = XZ², by Pythagorean theorem, it IS a right triangle.

Answer: yes

Question 7 · Triangle Classification & Circles

State the radius of the circle: x² + y² = 49

Show worked solution

The equation x² + y² = r² represents a circle centered at the origin.
Here, r² = 49, so r = 7.

Answer: 7

Question 8 · More Circles & Triangles

Classify the triangle with vertices A(-3,3), B(-1,5), and C(1,3). Also state whether it has a right angle.

Show worked solution

AB = √((-1+3)² + (5-3)²) = √(4 + 4) = √8
BC = √((1+1)² + (3-5)²) = √(4 + 4) = √8
AC = √((1+3)² + (3-3)²) = √16 = 4

Since AB = BC, the triangle is isosceles.
Check Pythagorean theorem: (√8)² + (√8)² = 8 + 8 = 16 = 4².
So it is also a right triangle.

Answer: isosceles right

Question 9 · More Midpoint & Distance

Find the length of the line segment with endpoints S(4,-5) and T(10,7). Round to the nearest tenth.

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Distance = √((10-4)² + (7-(-5))²) = √(6² + 12²) = √(36 + 144) = √180 ≈ 13.4

Answer: 13.4

Question 10 · More Midpoint & Distance

Find the length of the line segment with endpoints B(-6,8) and C(4,-5). Round to the nearest tenth.

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Distance = √((4-(-6))² + (-5-8)²) = √(10² + (-13)²) = √(100 + 169) = √269 ≈ 16.4

Answer: 16.4

Question 11 · Point-to-Line Shortest Distance Applications

Find the shortest distance from P(3,6)P(3, 6)P(3,6) to the line y=−23x+4y = -\frac{2}{3}x + 4y=−32​x+4. Round to the nearest tenth.

Show worked solution

The given slope is −23-\frac{2}{3}−32​, so the perpendicular slope is 32\frac{3}{2}23​.
Through P(3,6)P(3,6)P(3,6): y−6=32(x−3)y-6=\frac{3}{2}(x-3)y−6=23​(x−3), so y=32x+32y=\frac{3}{2}x+\frac{3}{2}y=23​x+23​.
Intersect with the original line:32x+32=−23x+4\frac{3}{2}x+\frac{3}{2}=-\frac{2}{3}x+423​x+23​=−32​x+4x=1513,y=4213x=\frac{15}{13},\quad y=\frac{42}{13}x=1315​,y=1342​.
Distance from PPP to this foot is(1513−3)2+(4213−6)2=121313≈3.3\sqrt{\left(\frac{15}{13}-3\right)^2+\left(\frac{42}{13}-6\right)^2}=\frac{12\sqrt{13}}{13}\approx3.3(1315​−3)2+(1342​−6)2​=131213​​≈3.3.
Therefore, the shortest distance is approximately 3.3 units.

Answer: 3.3

Question 12 · Point-to-Line Shortest Distance Applications

Find the shortest distance from P(2,−3)P(2, -3)P(2,−3) to the line y=4x−1y = 4x - 1y=4x−1. Round to the nearest tenth.

Show worked solution

The given slope is 4, so the perpendicular slope is −14-\frac14−41​.
Through P(2,−3)P(2,-3)P(2,−3): y+3=−14(x−2)y+3=-\frac14(x-2)y+3=−41​(x−2), so y=−14x−52y=-\frac14x-\frac52y=−41​x−25​.
Intersect with y=4x−1y=4x-1y=4x−1:4x−1=−14x−524x-1=-\frac14x-\frac524x−1=−41​x−25​x=−617,y=−4117x=-\frac6{17},\quad y=-\frac{41}{17}x=−176​,y=−1741​.
Distance from PPP to the foot is(−617−2)2+(−4117+3)2=101717≈2.4\sqrt{\left(-\frac6{17}-2\right)^2+\left(-\frac{41}{17}+3\right)^2}=\frac{10\sqrt{17}}{17}\approx2.4(−176​−2)2+(−1741​+3)2​=171017​​≈2.4.
Therefore, the shortest distance is approximately 2.4 units.

Answer: 2.4

Question 13 · More Midpoint & Distance

Find the length of the line segment with endpoints B(5,3) and C(1,-5). Round to the nearest tenth.

Show worked solution

Distance = √((1-5)² + (-5-3)²) = √((-4)² + (-8)²) = √(16 + 64) = √80 ≈ 8.9

Answer: 8.9

Common mistakes

Swapping x and y in the formulas.
Squaring wrong: (a−b)² ≠ a²−b².
Using the original slope instead of the negative reciprocal.
Rounding too early when an exact distance is asked.

Analytic Geometry Unit Test

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Ontario Grade 10 Analytic Geometry Practice Questions

What you need to know

Practice questions with solutions

Common mistakes

Analytic Geometry Unit Test

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