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.
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 (13)
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 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 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 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
A coastline is modeled by the line $$x + y = 10$$. A boat is at $$B(1, 2)$$. Find the shortest distance from the boat to the coastline as an exact simplified radical.
Show Solution
Line: $$x + y - 10 = 0$$. Slope of coastline is $$-1$$, so perpendicular slope is $$1$$.
Perpendicular through $$B$$: $$y - 2 = 1(x - 1) \Rightarrow y = x + 1$$.
Intersect with $$x + y = 10$$: $$x + (x + 1) = 10 \Rightarrow 2x = 9 \Rightarrow x = 4.5$$, $$y = 5.5$$.
$$d = \sqrt{(4.5-1)^2 + (5.5-2)^2} = \sqrt{3.5^2 + 3.5^2} = 3.5\sqrt{2} = \frac{7}{2}\sqrt{2}$$.
Therefore, the shortest distance is $$\dfrac{7\sqrt{2}}{2}$$.
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.
Show Solution
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.
Show Solution
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 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 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.
Find the shortest distance from $$P(3, 6)$$ to the line $$y = -\frac{2}{3}x + 4$$. Round to the nearest tenth.
Show Solution
The given slope is $$-\frac{2}{3}$$, so the perpendicular slope is $$\frac{3}{2}$$.
Through $$P(3,6)$$: $$y-6=\frac{3}{2}(x-3)$$, so $$y=\frac{3}{2}x+\frac{3}{2}$$.
Intersect with the original line:
$$\frac{3}{2}x+\frac{3}{2}=-\frac{2}{3}x+4$$
$$x=\frac{15}{13},\quad y=\frac{42}{13}$$.
Distance from $$P$$ to this foot is
$$\sqrt{\left(\frac{15}{13}-3\right)^2+\left(\frac{42}{13}-6\right)^2}=\frac{12\sqrt{13}}{13}\approx3.3$$.
Therefore, the shortest distance is approximately 3.3 units.
Find the shortest distance from $$P(2, -3)$$ to the line $$y = 4x - 1$$. Round to the nearest tenth.
Show Solution
The given slope is $$4$$, so the perpendicular slope is $$-\frac14$$.
Through $$P(2,-3)$$: $$y+3=-\frac14(x-2)$$, so $$y=-\frac14x-\frac52$$.
Intersect with $$y=4x-1$$:
$$4x-1=-\frac14x-\frac52$$
$$x=-\frac6{17},\quad y=-\frac{41}{17}$$.
Distance from $$P$$ to the foot is
$$\sqrt{\left(-\frac6{17}-2\right)^2+\left(-\frac{41}{17}+3\right)^2}=\frac{10\sqrt{17}}{17}\approx2.4$$.
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.