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MathsMaths509 views·Updated 31 Jul 2026·4 pages

Understanding Straight Lines: Simple Notes and Examples

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hana@hanajess777

Ever wondered how to prove three points lie on a...

1
of 4
straight lines  – page 1

Collinear Lines and Gradients

Collinear points all lie on the same straight line, and proving this is simpler than you might think. When points are collinear, all the line segments between them have identical gradients - that's your key to solving these problems.

To prove collinearity, calculate the gradients between different pairs of points using m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}. If they're equal, state that the segments are parallel and identify the common point they share.

Here's a handy tip: positive gradients slope upwards (like climbing a hill), whilst negative gradients slope downwards. The gradient actually equals tan θ, where θ is the angle between the line and the x-axis.

Quick Check: For negative slopes, find the angle using tan⁻¹, then subtract from 180° to get the actual angle with the x-axis.

2
of 4
straight lines  – page 2

Distance Formula and Perpendicular Lines

Finding the distance between two points is like using Pythagoras' theorem on a coordinate plane. The formula D=(x2x1)2+(y2y1)2D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} might look intimidating, but it's just creating a right-angled triangle and finding the hypotenuse.

Perpendicular gradients follow a simple rule: m1×m2=1m_1 \times m_2 = -1. To find a perpendicular gradient, flip the original fraction upside down and change the sign. If your gradient is 32\frac{3}{2}, the perpendicular gradient becomes 23-\frac{2}{3}.

When solving perpendicular line problems, rearrange the given equation into y=mx+cy = mx + c form first, then apply the perpendicular rule. Always leave distance answers in exact form (like 5\sqrt{5}) unless specifically asked to round.

Pro Tip: Remember that perpendicular lines meet at 90°, so their gradients are negative reciprocals of each other.

3
of 4
straight lines  – page 3

Medians and Altitudes of Triangles

A median connects the midpoint of one side to the opposite vertex - think of it as a line from the middle of a triangle's side to the far corner. Start by finding the midpoint using (x1+x22,y1+y22)\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right), then calculate the gradient to the opposite vertex.

An altitude is quite different - it drops perpendicularly from a vertex to the opposite side, creating a right angle. This means you'll need to find the gradient of the opposite side first, then use the perpendicular gradient rule.

Both problems follow the same final step: substitute your gradient and coordinates into yb=m(xa)y - b = m(x - a) to find the equation. The tricky bit is identifying which gradient and points to use for each type of line.

Memory Aid: Medians go to midpoints, altitudes are always perpendicular to the opposite side.

4
of 4
straight lines  – page 4

Perpendicular Bisectors

A perpendicular bisector combines everything you've learned - it passes through the midpoint of a line segment and meets it at a right angle. You're essentially creating both a median and an altitude in one go.

Follow this systematic approach: find the midpoint of the original line segment, calculate the gradient of that segment, then invert and change the sign to get your perpendicular gradient. Finally, use the equation yb=m(xa)y - b = m(x - a) with your midpoint coordinates.

These problems often appear in exam questions because they test multiple skills simultaneously. Take your time with the arithmetic - small errors in fraction calculations can throw off your entire answer.

Exam Strategy: Always double-check that your perpendicular gradient multiplies with the original gradient to give -1.

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That's right! Enjoy free access to study content, connect with fellow students, and get instant help – all at your fingertips.

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MathsMaths509 views·Updated 31 Jul 2026·4 pages

Understanding Straight Lines: Simple Notes and Examples

user profile picture
hana@hanajess777

Ever wondered how to prove three points lie on a straight line or find the exact distance between two coordinates? These coordinate geometry skills are essential for your Higher Maths exam and surprisingly useful in real life applications like computer...

1
of 4
straight lines  – page 1

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Collinear Lines and Gradients

Collinear points all lie on the same straight line, and proving this is simpler than you might think. When points are collinear, all the line segments between them have identical gradients - that's your key to solving these problems.

To prove collinearity, calculate the gradients between different pairs of points using m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}. If they're equal, state that the segments are parallel and identify the common point they share.

Here's a handy tip: positive gradients slope upwards (like climbing a hill), whilst negative gradients slope downwards. The gradient actually equals tan θ, where θ is the angle between the line and the x-axis.

Quick Check: For negative slopes, find the angle using tan⁻¹, then subtract from 180° to get the actual angle with the x-axis.

2
of 4
straight lines  – page 2

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Distance Formula and Perpendicular Lines

Finding the distance between two points is like using Pythagoras' theorem on a coordinate plane. The formula D=(x2x1)2+(y2y1)2D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} might look intimidating, but it's just creating a right-angled triangle and finding the hypotenuse.

Perpendicular gradients follow a simple rule: m1×m2=1m_1 \times m_2 = -1. To find a perpendicular gradient, flip the original fraction upside down and change the sign. If your gradient is 32\frac{3}{2}, the perpendicular gradient becomes 23-\frac{2}{3}.

When solving perpendicular line problems, rearrange the given equation into y=mx+cy = mx + c form first, then apply the perpendicular rule. Always leave distance answers in exact form (like 5\sqrt{5}) unless specifically asked to round.

Pro Tip: Remember that perpendicular lines meet at 90°, so their gradients are negative reciprocals of each other.

3
of 4
straight lines  – page 3

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  • Access to all documents
  • Improve your grades
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Medians and Altitudes of Triangles

A median connects the midpoint of one side to the opposite vertex - think of it as a line from the middle of a triangle's side to the far corner. Start by finding the midpoint using (x1+x22,y1+y22)\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right), then calculate the gradient to the opposite vertex.

An altitude is quite different - it drops perpendicularly from a vertex to the opposite side, creating a right angle. This means you'll need to find the gradient of the opposite side first, then use the perpendicular gradient rule.

Both problems follow the same final step: substitute your gradient and coordinates into yb=m(xa)y - b = m(x - a) to find the equation. The tricky bit is identifying which gradient and points to use for each type of line.

Memory Aid: Medians go to midpoints, altitudes are always perpendicular to the opposite side.

4
of 4
straight lines  – page 4

Sign up to see the content. It's free!

  • Access to all documents
  • Improve your grades
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By signing up you accept Terms of Service and Privacy Policy

Perpendicular Bisectors

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Follow this systematic approach: find the midpoint of the original line segment, calculate the gradient of that segment, then invert and change the sign to get your perpendicular gradient. Finally, use the equation yb=m(xa)y - b = m(x - a) with your midpoint coordinates.

These problems often appear in exam questions because they test multiple skills simultaneously. Take your time with the arithmetic - small errors in fraction calculations can throw off your entire answer.

Exam Strategy: Always double-check that your perpendicular gradient multiplies with the original gradient to give -1.

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Students love us — and so will you.

4.6/5App Store
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The app is very easy to use and well designed. I have found everything I was looking for so far and have been able to learn a lot from the presentations! I will definitely use the app for a class assignment! And of course it also helps a lot as an inspiration.

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Wow, I am really amazed. I just tried the app because I've seen it advertised many times and was absolutely stunned. This app is THE HELP you want for school and above all, it offers so many things, such as workouts and fact sheets, which have been VERY helpful to me personally.

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