Straight Lines Graphs – SS2 Mathematics Lesson Note (Second Term)

A straight line graph is the graph of a linear equation of the form y = mx + c or ax + by + c = 0. The gradient of the line, which is the change in y divided by the change in x, is the same at every point on the line.

Behavioural objectives

By the end of the lesson, students should be able to:

• write the coordinates of a point in the correct order;
• find the point where two lines meet, and where a line meets the x-axis or the y-axis;
• calculate the gradient of a straight line and relate it to the tangent of the angle of slope;
• write the equation of a straight line from given information;
• explain how the gradient of a curve at a point is found from a tangent.

What are the coordinates of a point?

In coordinate geometry, we make use of points in a plane. A point consists of the x-coordinate, called the abscissa, and the y-coordinate, known as the ordinate.

When locating a point on the x-y plane, the x-coordinate is written first and then the y-coordinate. For example, in a given point (a, b), the value of x is a and that of y is b. Similarly, in the point (3, 5), the value of x is 3 and that of y is 5.

What is a straight line graph?

A linear equation gives a straight line graph. The equation is in the general form y = mx + c or ax + by + c = 0.

Example: Draw the graph of the equation 4x + 2y = 5.

How do you find where lines meet?

Point of intersection of two linear equations

Take two lines y = ax + b and y = cx + d. They intersect where ax + b = cx + d. That is, you solve the two equations simultaneously.

Example: Find the point of intersection of the lines y = 3x + 2 and y = 2x + 5.

Solution

y = 3x + 2 (1)
y = 2x + 5 (2)

At the point of intersection:

3x + 2 = 2x + 5
3x – 2x = 5 – 2
x = 3

Substitute 3 for x in equation (1): y = 3(3) + 2 = 11.

Hence, the point of intersection is (3, 11).

Intersection of a line with the x-axis or y-axis

The point where a line meets the x-axis is found by putting y = 0 to find the corresponding value of x. If that value is a, the required point is (a, 0).

Similarly, for the point where a line meets the y-axis, put x = 0 to find the corresponding value of y. If that value is b, the required point is (0, b).

Example: Find the point of intersection of the line 2x + 3y + 2 = 0 with the (i) x-axis (ii) y-axis.

What is the gradient of a straight line?

The gradient of a straight line is defined as the ratio of the change in y to the change in x in moving from one point to another on the line. The gradient of a straight line is always constant.

This means that the gradient of the line is the ratio of the increase in y to the increase in x.

Tangent of the angle of slope

In the diagram above, let θ be the angle of slope. The change in y is y₂ – y₁ and the change in x is x₂ – x₁. So:

tan θ = (y₂ – y₁) / (x₂ – x₁) = m

It follows that the gradient of a line can be defined as the tangent of the angle of slope.

Example: Calculate the gradient and the angle of slope of the line passing through (1, 3) and (-4, 2).

How do you find the equation of a straight line?

Line with gradient m and y-intercept c

The equation of a line with gradient m and y-intercept c is y = mx + c.

Line passing through a point (x₁, y₁) with a given gradient

The general equation of a line with known gradient m, which passes through the point (x₁, y₁), is given by:

m = (y – y₁) / (x – x₁)

Example: Find the equation of the line with gradient 2 which passes through the point (-3, 2).

Solution

Here, m = 2 and (x₁, y₁) = (-3, 2). The required equation of the line is:

(y – 2) / (x – (-3)) = 2
(y – 2) / (x + 3) = 2
y – 2 = 2(x + 3)
y = 2x + 6 + 2
y = 2x + 8

Line passing through two given points

The equation of a line passing through two given points (x₁, y₁) and (x₂, y₂) is:

(y – y₁) / (x – x₁) = (y₂ – y₁) / (x₂ – x₁)

Double intercept form

The equation of a line which has an intercept a on the x-axis and an intercept b on the y-axis is given by:

x/a + y/b = 1

Line passing through a point and making an angle θ with the horizontal axis

The equation of a line passing through the point (x₁, y₁) and making an angle θ with the horizontal axis is:

(y – y₁) / (x – x₁) = tan θ, or y – y₁ = (x – x₁) tan θ

How do you draw a tangent to a curve?

The gradient at any particular point on a curve is the gradient of the tangent to the curve at that point. For example, the gradient of the curve at point A is the gradient of the tangent BA, that is, tan θ.

The tangent is drawn by placing a ruler against the curve at A and drawing a line so that the angles between the line and the curve are equal.

Note: At a turning point (a maximum or minimum), the tangent is horizontal, so the gradient of the curve there is zero.

Key points

• The gradient of a straight line is the change in y divided by the change in x.
• The gradient of a curve at a point is the gradient of the tangent at that point.
• The gradient at the turning point of any quadratic equation equals zero.

Evaluation

1. State the general forms of the equation of a straight line.
2. Find the point of intersection of the line 2x + 3y + 2 = 0 with the x-axis and with the y-axis.
3. Calculate the gradient and the angle of slope of the line passing through (1, 3) and (-4, 2).
4. Find the equation of the line with gradient 2 which passes through the point (-3, 2).
5. What is the gradient of a curve at its turning point, and why?

Assignment

Exercise 14.5, number 1 (from the recommended text). The figure below represents the graph of the function y = x² + 4x – 5.

1. (a) Use the given tangents to find the gradient of the curve at (i) A (ii) B.
2. (b) Use the graph to find the roots of the function.
3. (c) State the equation of the line of symmetry of the curve.