Unit notes and practice questions

Grade 10 Mathematics unit 7: Coordinate Geometry

Everything here is for unit 7 of the Grade 10 Mathematics textbook: what the chapter covers, the short notes, practice questions with their answers, and the lesson video where one exists.

5
Cards: 10
Practice questions on unit 7

What unit 7 covers

1.Distance between two points
2.Division of a line segment
3.Equation of a line
4.Slopes of parallel and perpendicular lines
5.Applications

Short notes for unit 7

The cards filed under this chapter on the Grade 10 Mathematics page. Open one at a time.

Area of a Triangle in Coordinate PlaneFormula

A = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|

A = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|

$A$: Area of the triangle $\text{square units}$
$x_1, y_1$: Coordinates of the first vertex $\text{units}$
$x_2, y_2$: Coordinates of the second vertex $\text{units}$
$x_3, y_3$: Coordinates of the third vertex $\text{units}$

When you use it

Use this formula to calculate the area of a triangle when the coordinates of its three vertices are given.

Watch out

Always apply the absolute value at the end. Area can never be negative, and if it evaluates to zero the three points are collinear.

Drafted from Grade 10 Mathematics, page 328 onwards, then checked twice before it went up

Distance FormulaFormula

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

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

$d$: Distance between two points $\text{units}$
$x_1, y_1$: Coordinates of the first point $\text{units}$
$x_2, y_2$: Coordinates of the second point $\text{units}$

When you use it

Use this formula to compute the straight line distance between any two points in the Cartesian plane.

Watch out

Do not mix coordinates from different points. Subtract x from x, and y from y, before squaring.

Drafted from Grade 10 Mathematics, page 328 onwards, then checked twice before it went up

Parallel and Perpendicular LinesReference tableParallel lines · Perpendicular lines

Line Relationship	Slope Condition	Geometric Meaning
Parallel lines	$m_1 = m_2$	Equal inclination angles
Perpendicular lines	$m_1 \cdot m_2 = -1$	Intersect at right angle

When you use it

Use these conditions to check whether two non-vertical lines are parallel or perpendicular, or to construct perpendicular and parallel line equations.

Watch out

The perpendicular condition only applies to non-vertical and non-horizontal lines. A vertical line is perpendicular to a horizontal line.

Drafted from Grade 10 Mathematics, page 328 onwards, then checked twice before it went up

Section FormulaFormula

(x_0, y_0) = \left(\frac{p x_2 + q x_1}{p + q}, \frac{p y_2 + q y_1}{p + q}\right)

(x_0, y_0) = \left(\frac{p x_2 + q x_1}{p + q}, \frac{p y_2 + q y_1}{p + q}\right)

$x_0, y_0$: Coordinates of the dividing point $\text{units}$
$p : q$: Ratio of internal division $\text{dimensionless}$
$x_1, y_1$: Coordinates of the first endpoint $\text{units}$
$x_2, y_2$: Coordinates of the second endpoint $\text{units}$

When you use it

Use this formula to find the coordinates of a point dividing a line segment internally in a given ratio p:q.

Watch out

Multiply the first ratio part p with the second point coordinates, and the second ratio part q with the first point coordinates.

Drafted from Grade 10 Mathematics, page 328 onwards, then checked twice before it went up

Slope and InclinationFormula

m = \frac{y_2 - y_1}{x_2 - x_1} = \tan\theta

m = \frac{y_2 - y_1}{x_2 - x_1} = \tan\theta

$m$: Slope of the line $\text{dimensionless}$
$\theta$: Angle of inclination $^{\circ}$
$x_1, y_1$: Coordinates of the first point $\text{units}$
$x_2, y_2$: Coordinates of the second point $\text{units}$

When you use it

Use this formula to compute the gradient of a straight line from two known points or from its angle of inclination with the positive x axis.

Watch out

A vertical line where x1 equals x2 has an undefined slope, not zero slope. A horizontal line has slope zero.

Drafted from Grade 10 Mathematics, page 328 onwards, then checked twice before it went up

Practice questions on unit 7

10 multiple-choice questions written against this chapter, each with its answer and the reason. Try one before you open the answer.

Question 1

Find the distance between the points C(2, 3) and D(6, 0).

A.5 units
B.7 units
C.25 units
D.1 unit

Show the answer and why

Answer: A 5 units

Why

Use the distance formula d = √((x₂ − x₁)² + (y₂ − y₁)²).

Here x₂ − x₁ = 6 − 2 = 4 and y₂ − y₁ = 0 − 3 = −3.

d = √(4² + (−3)²) = √(16 + 9) = √25 = 5 units.

Question 2

Find the midpoint of the line segment joining the points (2, 5) and (8, 1).

A.(5, 3)
B.(3, 2)
C.(10, 6)
D.(6, 3)

Show the answer and why

Answer: A (5, 3)

Why

The midpoint formula is ((x₁ + x₂)/2, (y₁ + y₂)/2).

x = (2 + 8)/2 = 5 and y = (5 + 1)/2 = 3. So the midpoint is (5, 3).

Question 3

Find the gradient (slope) of the line passing through the points (2, 3) and (6, 11).

A.2
B.1/2
C.−2
D.4

Show the answer and why

Answer: A 2

Why

Gradient m = (y₂ − y₁)/(x₂ − x₁) = (11 − 3)/(6 − 2) = 8/4 = 2.

Question 4

A line passes through the points (5, −2) and (5, 8). What is its gradient?

A.0
B.1
C.No gradient, because the line is vertical
D.−1

Show the answer and why

Answer: C No gradient, because the line is vertical

Why

Both points have x = 5, so the x-coordinates are equal. The line is vertical (parallel to the y-axis). A vertical line has no slope, so it has no gradient.

Question 5

A point R divides the line segment joining A(1, 2) and B(4, 5) internally in the ratio 1 : 2. Find the coordinates of R.

A.(2, 3)
B.(3, 4)
C.(1, 2)
D.(3, 2)

Show the answer and why

Answer: A (2, 3)

Why

Use the section formula with p = 1 and q = 2, where P(x₁, y₁) = (1, 2) and Q(x₂, y₂) = (4, 5).

x = (p·x₂ + q·x₁)/(p + q) = (1·4 + 2·1)/3 = 6/3 = 2.

y = (p·y₂ + q·y₁)/(p + q) = (1·5 + 2·2)/3 = 9/3 = 3.

So R = (2, 3).

Question 6

A straight line passes through the points (0, 5) and (2, 9). Which equation describes this line?

A.y = 2x + 5
B.y = 5x + 2
C.y = 2x − 5
D.y = −2x + 5

Show the answer and why

Answer: A y = 2x + 5

Why

First find the slope: m = (9 − 5)/(2 − 0) = 4/2 = 2.

Use the slope-intercept form y = mx + b. The point (0, 5) gives b = 5, because x = 0 makes y = b.

So the equation is y = 2x + 5.

Question 7

Find the equation of the line that passes through the point (2, 1) and is parallel to the line y = 3x − 4.

A.y = 3x − 5
B.y = −3x + 7
C.y = 3x + 1
D.y = x/3 − 5

Show the answer and why

Answer: A y = 3x − 5

Why

Parallel lines have the same slope, so m = 3 (the slope of y = 3x − 4).

Use the point-slope form y − y₁ = m(x − x₁) with (x₁, y₁) = (2, 1):

y − 1 = 3(x − 2), so y − 1 = 3x − 6, and y = 3x − 5.

Question 8

A line has gradient 2/3. What is the gradient of a line that is perpendicular to it?

A.−3/2
B.3/2
C.−2/3
D.2/3

Show the answer and why

Answer: A −3/2

Why

Two non-vertical lines are perpendicular when m₁ × m₂ = −1.

So m₂ = −1 ÷ (2/3) = −1 × (3/2) = −3/2.

Question 9

Find the area of the triangle with vertices (2, 3), (0, −1) and (3, −5).

A.10 square units
B.20 square units
C.5 square units
D.12 square units

Show the answer and why

Answer: A 10 square units

Why

Use A = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|.

Substitute (x₁, y₁) = (2, 3), (x₂, y₂) = (0, −1), (x₃, y₃) = (3, −5):

2(−1 + 5) + 0(−5 − 3) + 3(3 + 1) = 8 + 0 + 12 = 20.

A = ½ × 20 = 10 square units.

Question 10

In a city grid, each unit is 1 km. A bus stop is at point A(−2, 3) and a school is at point B(4, −5). How far is the school from the bus stop in a straight line?