Comprehensive Higher Maths Study Notes

Equation of a Straight Line

Ever wondered how mathematicians describe lines precisely? You need just two pieces of information: a gradient and a point.

The gradient formula shows how steep a line is: m = $y₂ - y₁$/$x₂ - x₁$. This measures the vertical change divided by the horizontal change between two points.

For example, to find the gradient of a line passing through A$-3, 5$ and B(1, 7): m = $7 - 5$/$1 - (-3)$ = 2/4 = 1/2

Every straight line can be written in the form y = mx + c, where m is the gradient and c is the y-intercept (where the line crosses the y-axis).

💡 When finding the gradient from an equation that's not in this form, always rearrange to isolate y on one side. For example, with 3x + 2y - 5 = 0, we rearrange to y = -3x/2 + 5/2, giving a gradient of -3/2.

Finding Line Equations

There are two main formulas for finding the equation of a straight line:

1. y = mx + c (when you know the gradient and y-intercept)
2. y - b = m$x - a$ (when you know the gradient and a point (a,b))

Using the second formula is particularly handy when you have a point and a gradient. For example, if we have a line through (4,5) with gradient 2: y - 5 = 2$x - 4$ y - 5 = 2x - 8 y = 2x - 3

When finding the equation of a line joining two points, first calculate the gradient using the gradient formula, then use either point with the point-gradient formula.

For example, with points $-1,-2$ and (3,10):

1. Calculate gradient: m = $10-(-2)$/$3-(-1)$ = 12/4 = 3
2. Use either point with y - b = m$x - a$: y - 10 = 3$x - 3$ y = 3x - 1

💡 Always double-check your work by substituting one of your points into your final equation to verify it works!

More Line Equations and Collinearity

Let's continue with finding line equations. For a line through $-5,-1$ with gradient -2/3: y + 1 = -2/3$$x + 5 Multiply both sides by 3: 3y + 3 = -2$x + 5$ Expand: 3y + 3 = -2x - 10 Rearrange: 3y = -2x - 13

Collinear points lie on the same straight line. To check if three points are collinear, calculate the gradient between each pair of points – if they're equal, the points are collinear.

For example, to prove D$-1,5$, E(0,2), and F$4,-10$ are collinear:

1. Calculate mₐₑ = $2-5$/$0-(-1)$ = -3/1 = -3
2. Calculate mₑₖ = $-10-2$/$4-0$ = -12/4 = -3

Since mₐₑ = mₑₖ = -3, and E is a common point, we've proven that D, E, and F lie on the same straight line.

💡 Collinearity can also be checked by showing that the three points satisfy the same straight line equation.

Angles Between Lines and Axes

The gradient of a line can tell us the angle it makes with the positive x-axis using the relationship: m = tan θ.

To find this angle, we use the inverse tangent function: θ = tan⁻¹$m$.

For example, with y = x - 1:

• Gradient m = 1
• θ = tan⁻¹(1) = 45°

With y = 5 - √3x:

• Gradient m = -√3
• θ = tan⁻¹$-√3$ = 120°

When dealing with points, first calculate the gradient. For a line joining $3,-2$ and (1,4):

• m = $4-(-2)$/$1-3$ = 6/$-2$ = -3
• θ = tan⁻¹$-3$ = 56.3°

💡 When working with negative gradients, remember that angles measured from the positive x-axis can be expressed as either the acute angle in the fourth quadrant or the equivalent angle in the second quadrant (180° - acute angle).

The Distance Formula

To find the distance between two points, we use Pythagoras' theorem in coordinate form: d = √$(x₂ - x₁)² + (y₂ - y₁)²$

For example, to find the distance between A$-1,1$ and B$2,-1$: d = √$(2-(-1))² + (-1-1)²$ d = √$3² + (-2)²$ d = √$9 + 4$ d = √13

For points Q(11,2) and R$-2,-5$: d = √$(-2-11)² + (-5-2)²$ d = √$(-13)² + (-7)²$ d = √$169 + 49$ d = √218 ≈ 14.8

The distance formula is extremely useful for comparing lengths. For example, to show that BC = 2AB for points A$2,-1$, B$5,-2$ and C(7,4), we calculate:

• AB = √$(5-2)² + (-2-(-1))²$ = √$9 + 1$ = √10
• BC = √$(7-5)² + (4-(-2))²$ = √$4 + 36$ = √40 = 2√10

💡 When simplifying square roots, look for perfect square factors. For example, √40 = √(4×10) = 2√10.

Perpendicular Lines

Two lines are perpendicular when they meet at a 90° angle. The key property of perpendicular lines is that their gradients multiply to give -1: m₁ × m₂ = -1

This means if a line has gradient m, then any perpendicular line to it has gradient -1/m.

For example:

• If m = 3/2, then m₁ = -2/3 (perpendicular gradient)
• If m = -1.5, then m₁ = 2/3 (perpendicular gradient)

To find a perpendicular gradient for a line not in the form y = mx + c, first rearrange to find m.

For the line 2y + 5x = 0:

• Rearrange: 2y = -5x, so y = -5x/2
• Gradient m = -5/2
• Perpendicular gradient m₁ = 2/5

💡 This perpendicular property is extremely useful in geometry problems involving right angles, such as finding altitudes and perpendicular bisectors of triangles.

Perpendicular Lines and Median

Continuing with perpendicular lines, let's consider more examples:

If a line has equation y = 3x/2 + 4:

• Gradient m = 3/2
• Perpendicular gradient m₁ = -2/3

Remember that perpendicular lines always satisfy the relationship m₁ × m₂ = -1.

A median of a triangle is a line that joins a vertex to the midpoint of the opposite side. Every triangle has three medians, and they always meet at a single point called the centroid.

To find the equation of a median:

1. Find the midpoint of the side
2. Calculate the gradient between the vertex and this midpoint
3. Use y - b = m$x - a$ to find the line equation

For example, in a triangle with vertices P(0,2), Q(4,4), and R$8,-6$, to find the median through P:

1. Midpoint of QR = $(4+8)/2, (4+(-6))/2$ = $6,-1$
2. Gradient from P to midpoint = $-1-2$/$6-0$ = -3/6 = -1/2
3. Equation: y - 2 = -1/2$x - 0$, which simplifies to y = -x/2 + 2

💡 Medians are important in physics for finding the center of mass of triangular objects.

Finding Medians in Triangles

Let's continue working with the median of the triangle with vertices P(0,2), Q(4,4), and R$8,-6$.

We've calculated that the midpoint of QR is $6,-1$ and the gradient from P to this midpoint is -1/2.

Using the point-gradient formula with P(0,2): y - 2 = -1/2$x - 0$ y - 2 = -x/2 y = -x/2 + 2

This equation represents the median from vertex P to the midpoint of side QR.

The steps for calculating any median are:

1. Calculate the midpoint of the side opposite to your chosen vertex
2. Find the gradient between the vertex and this midpoint
3. Use the point-gradient form to write the equation

When sketching these lines, it helps to plot the vertex and the midpoint, then draw the line connecting them. The resulting median divides the triangle into two parts with equal areas.

💡 The three medians of a triangle always intersect at a single point called the centroid, which is located 2/3 of the way from any vertex to the midpoint of the opposite side.

Altitude and Perpendicular Bisector

An altitude of a triangle is a line through a vertex that is perpendicular to the opposite side.

To find the equation of an altitude:

1. Calculate the gradient of the side $m$
2. Find the perpendicular gradient $-1/m$
3. Use the point-gradient formula with the vertex

For our triangle with vertices P(0,2), Q(4,4), and R$8,-6$:

• The gradient of PQ is $4-2$/$4-0$ = 2/4 = 1/2
• The perpendicular gradient is -2
• Using point R$8,-6$: y - $-6$ = -2$x - 8$
• Simplified: y + 6 = -2x + 16, or y = -2x + 10

A perpendicular bisector of a line segment:

• Passes through the midpoint of the segment
• Is perpendicular to the segment

These properties make perpendicular bisectors essential for finding points that are equidistant from two given points.

💡 The three altitudes of a triangle are concurrent, meaning they all pass through a single point called the orthocenter.

Perpendicular Bisector

To find the equation of a perpendicular bisector of a line segment PQ:

1. Calculate the midpoint of PQ
2. Find the gradient of PQ
3. Calculate the perpendicular gradient
4. Use the point-gradient formula with the midpoint

For example, with P$3,-2$ and Q$-5,4$:

1. Midpoint = $(3+(-5))/2, (-2+4)/2$ = $-1,1$
2. Gradient of PQ = $4-(-2)$/$-5-3$ = 6/$-8$ = -3/4
3. Perpendicular gradient = 4/3
4. Equation: y - 1 = 4/3$x + 1$

Multiplying both sides by 3: 3y - 3 = 4x + 4 3y = 4x + 7

Perpendicular bisectors have a special property: any point on a perpendicular bisector is equidistant from the two endpoints of the original line segment.

This property makes perpendicular bisectors crucial for constructing the circumcenter of a triangle (the center of the circle that passes through all three vertices).

💡 The perpendicular bisectors of the three sides of a triangle all intersect at a single point, which is the center of the circumscribed circle of the triangle.