Notes · Unit 8
Geometry of Straight Lines
Fourteen short slides on gradient and inclination, the different forms of a line's equation, parallel and perpendicular lines, angles between lines and intersections.
Press Next (or the right arrow key) to move on. On most slides, each press reveals one more idea.
GradientForms of equationsParallel and perpendicularAngle between linesIntersections
1 · Inclination and gradient
How steep is a line?
Inclination θ is the angle the line makes with the positive x-axis, measured anticlockwise.
Gradient (slope). m = tan θ. For example, inclination 45° gives m = tan 45° = 1.
Horizontal line: θ = 0°, so m = 0.
Vertical line: θ = 90°, so the run is zero and the gradient is undefined.
Vertical line: θ = 90°, so the run is zero and the gradient is undefined.
Sign. A line rising from left to right has m > 0 (0° < θ < 90°). A negative gradient means 90° < θ < 180°.
2 · Gradient from two points
Rise over run
m = (y2 − y1)/(x2 − x1)
(2, 6) and (5, 8): m = (8 − 6)/(5 − 2) = 2/3
Either point can come first, as long as both differences are taken in the same order.
Do not mix the order. For (1, 2) and (3, 6), writing (1 − 3)/(6 − 2) is wrong. The correct value is (6 − 2)/(3 − 1) = 2.
Real life. A hill with gradient 1/25 rises 1 unit for every 25 units of horizontal distance.
3 · Parallel and perpendicular
Compare the gradients
Parallel
Equal gradients: m1 = m2. They make the same angle with the x-axis.
Perpendicular
Product −1: m1 × m2 = −1.
Examples. 2/3 × (−3/2) = −1, so these slopes are perpendicular. So are 1 and −1.
Slip. The line perpendicular to slope 2 does not have slope −2, because 2 × (−2) = −4. It has slope −1/2.
Collinear points. If the slopes of AB and BC are equal, the lines share point B and are the same line. For (1, 1), (2, 2), (3, 4) the slopes are 1 and 2, so the points are not collinear.
4 · Equations of a line
Horizontal, vertical and slope forms
Vertical line: x = a, parallel to the y-axis. Horizontal line: y = b, parallel to the x-axis. The horizontal line through (2, 3) is y = 3. The x-axis is y = 0.
Slope-intercept form: y = mx + c, where c is the y-intercept. For y = −2x + 5: m = −2 and c = 5.
Point-slope form: y − y1 = m(x − x1), for a line through (x1, y1) with slope m.
Point-slope form cannot describe a vertical line, because its slope is undefined. Use x = a.
5 · Equations of a line
Two-point, intercept and normal forms
Two-point form. Through (2, −1) and (5, 8): slope = 9/3 = 3, so y + 1 = 3(x − 2), which gives y = 3x − 7.
Two-intercept form. x/a + y/b = 1 for a line through (a, 0) and (0, b). With intercepts 3 and 5: x/3 + y/5 = 1, and multiplying by 15 gives 5x + 3y = 15.
Normal form. x cos α + y sin α = p, where p is the perpendicular distance from the origin to the line, so p is positive.
6 · The general form
ax + by + c = 0
a and b must not both be zero. If both were, the equation would become c = 0, which is not a line.
Reading it off (b ≠ 0): slope −a/b, y-intercept −c/b. For a ≠ 0 the x-intercept is −c/a.
3x − 4y + 8 = 0: slope = −3/(−4) = 3/4.
2x + 3y − 6 = 0: the slope is −2/3, not 2/3. Writing y = −(2/3)x + 2 shows it.
If b = 0, as in 5x − 6 = 0, then x = −c/a, a vertical line. A line through the origin has no constant term, like y = 3x.
7 · Worked examples
Use a point and a condition
Perpendicular line. Find the line through (2, −3) perpendicular to a line of slope 2/3.
The slope is −3/2. So y + 3 = −(3/2)(x − 2), giving 2y + 6 = −3x + 6, that is 3x + 2y = 0.
The slope is −3/2. So y + 3 = −(3/2)(x − 2), giving 2y + 6 = −3x + 6, that is 3x + 2y = 0.
Finding a constant. The line 3x + ky − 11 = 0 passes through (1, 2). Substitute: 3 + 2k − 11 = 0, so k = 4.
Substituting a point into an equation tests whether the point lies on the line.
8 · Angle between two lines
From slope m1 to slope m2
tan θ = (m2 − m1)/(1 + m1m2)
Slopes 1/2 and 3: tan θ = (3 − 1/2)/(1 + 3/2) = (5/2)/(5/2) = 1, so θ = 45°.
Slopes 0 and √3: tan θ = √3/1, so θ = 60°.
Parallel lines: m1 = m2 makes the top zero, so θ = 0°.
Perpendicular lines: m1m2 = −1 makes the denominator zero, so tan θ is undefined and θ = 90°.
9 · Intersecting lines
Where two lines meet
Two lines with different slopes meet at exactly one point.
x + y = 7 and x − y = 1. Adding gives 2x = 8, so x = 4 and y = 3. The point is (4, 3).
Family of lines. Every line of the form (a1x + b1y + c1) + k(a2x + b2y + c2) = 0 passes through the intersection point of the two original lines, because that point makes both brackets zero.
10 · Real-life lines
Slope is a rate, intercept is a start
Taxi. C = 25d + 100. The slope 25 is the cost per kilometre. The y-intercept 100 is the fixed charge at d = 0.
Temperature. F = (9/5)C + 32. The slope 9/5 is the change in F for each 1° change in C.
Meter reading. It shows 50 units and 12 units are used per day. 50 + 12d = 410 gives 12d = 360, so after 30 days.
11 · Common mistakes
Five traps to avoid
Perpendicular slope. It is the negative reciprocal, not just the negative.
Wrong sign for the slope of ax + by + c = 0. It is −a/b.
Reversing the order in one difference only in the gradient formula.
Giving a vertical line a gradient. Its gradient is undefined, and its equation is x = a.
Forgetting what equal slopes mean. Two lines with equal slopes are parallel. If they also share a point, they are the same line.
12 · Quick check
Try these, then tap to check
13 · Your turn
Practise Unit 8
Three timed MCQ tests with an explanation after every answer.
Practice questions are independently written and are not official FBISE papers.