A-level Mathematics/CIE/Pure Mathematics 1/Coordinate Geometry

The Equation of a Line

Calculations involving lines

Distance between two points

The distance between two points can be found with the Pythagorean theorem

The distance between two points is given by the formula $Dist={\sqrt {\Delta x^{2}+\Delta y^{2}}}$ where $\Delta x$ is the difference in x-values between the two points and $\Delta y$ is the difference in y-values between the two points.

This formula can also be seen as applying the Pythagorean theorem to the points, where the differences in x-values and y-values form two sides of a right-angled triangle.

Midpoint of two points

The midpoint is the point halfway between two points.

The midpoint is the point which is exactly halfway between two points. The coordinates of the midpoint are given by $({\frac {x_{1}+x_{2}}{2}},{\frac {y_{1}+y_{2}}{2}})$ where $(x_{1},y_{1})$ and $(x_{2},y_{2})$ are the coordinates of the two points.

e.g. The midpoint between $(2,-1)$ and $(4,3)$ is $(3,1)$

You may notice that this expression states that the midpoint's x-coordinate is the average of the x-coordinates of the points, and its y-coordinate is the average of the points' y-coordinates. Essentially, this means that the midpoint is the average of the two points.

Gradients

The gradient $m$ of a line is determined by the ratio $m={\frac {rise}{run}}$ where $rise$ is the change in y-value and $run$ is the change in x-value.

This can also be expressed as $m={\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}$ when finding the gradient of a line between two points.

e.g. The gradient of a line that passes through $(2,4)$ and $(3,6)$ has gradient ${\frac {6-4}{3-2}}={\frac {2}{1}}=2$

Intersecting lines

An intersection between two lines

When two lines intersect, the point of intersection is where the two lines cross. The point of intersection is thus on both lines, meaning that it can be found using simultaneous equations.

e.g. The lines $y=2x+3$ and $y=-x$ intersect. Find the point of intersection.

${\begin{aligned}y&=2x+3\\y&=-x\\-x&=2x+3\\-3x&=3\\x&=-1\\y&=-(-1)=1\\\therefore {\text{The}}&{\text{ point of intersection is at }}(-1,1)\end{aligned}}$

Parallel lines

The lines AB and CD are parallel.

Parallel lines always have the same gradient, and do not intersect.

e.g. The lines $y=3x+1$ and $y=3x-2$ are parallel.

Sometimes we'll need to find a line which is parallel to a given line and passes through a given point.

e.g. Find the equation of a line parallel to $y=2x+3$ that passes through $(5,6)$

${\begin{aligned}Parallel\implies &y=2x+c\\(5,6)\implies &6=2(5)+c\\&6=10+c\\&c=-4\\\therefore \quad &y=2x-4\end{aligned}}$

Perpendicular lines

The blue line and red line are perpendicular

Perpendicular lines are at right angles to one another. The product of the gradients of two perpendicular lines is always -1.

e.g. The lines $y=x$ and $y=-x$ are perpendicular.

Sometimes we'll need to find a perpendicular line that goes through a specific point.

e.g. Find the equation of the line perpendicular to $y=-{\frac {x}{2}}+3$ that passes through the origin.

${\begin{aligned}Perpendicular\implies &y={\frac {-1}{-1/2}}x+c\\&y=2x+c\\Through\ origin\implies &0=2(0)+c\\&c=0\\\therefore \quad &y=2x\end{aligned}}$

Different forms of the equation of a line

There are three main ways to write an equation of a line:

• $y=mx+c$
• ${\frac {y-y_{1}}{x-x_{1}}}=m$
• $ax+by+c=0$

Finding the equation of a line from a point and the gradient

The equation of a line can be found using a point and the gradient using the second equation ${\frac {y-y_{1}}{x-x_{1}}}=m$ followed by rearranging the equation to the form $y=mx+c$.

e.g. A line with gradient $2$ goes through the point $(2,3)$. Find its equation.

${\begin{aligned}{\frac {y-3}{x-2}}&=2\\y-3&=2(x-2)\\y&=2x-4+3\\y&=2x-1\end{aligned}}$

Finding the equation of a line from two points

When given two points, we can find the gradient using $m={\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}$. Using this gradient, the same method can be used as for a point and the gradient.

e.g. A line travels through the points $(2,-3)$ and $(1,1)$. Find the equation of the line.

${\begin{aligned}m={\frac {1+3}{1-2}}&={\frac {4}{-1}}=4\\{\frac {y+3}{x-2}}&=4\\y+3&=4(x-2)\\y+3&=4x-8\\y&=4x-11\end{aligned}}$

The Equation of a Circle

A circle consists of all points that are a given distance from its centre. The distance between two points can be defined using the Pythagorean theorem $a^{2}+b^{2}=c^{2}$. Thus, the equation of a circle centred at the origin is given by $x^{2}+y^{2}=r^{2}$ where $r$ is the radius of the circle.

e.g. A circle centred at the origin with radius $5$ would have the equation $x^{2}+y^{2}=25$

If the circle is not centred at the origin, we can translate this equation to a different point. Thus, the equation becomes $(x-x_{c})^{2}+(y-y_{c})^{2}=r^{2}$ where $(x_{c},y_{c})$ are the coordinates of the centre.

e.g. A circle centred at $(1,2)$ with radius ${\sqrt {5}}$ would have the equation $(x-1)^{2}+(y-2)^{2}=5$

Interactions between Lines & Circles

When given a problem where a line and a circle intersect, it is useful to use a substitution method of solving simultaneous equations.

e.g. The line $y=1-x$ intersects the circle $x^{2}+y^{2}=1$. Find the coordinates of these intersection points

${\begin{aligned}x^{2}+y^{2}&=1\\y&=1-x\\x^{2}+(1-x)^{2}&=1\\x^{2}+1-2x+x^{2}&=1\\2x^{2}-2x&=0\\2x(x-1)&=0\\x&=\{0,1\}\\y&=\{1,0\}\\\therefore &\ {\text{Intersections are at (0,1) and (1,0)}}\end{aligned}}$

Interactions between Lines & Quadratics

When a quadratic and a line intersect, we can again use substitution to find the points of intersection.

e.g. The line $y=3x-8$ intersects the quadratic $y=x^{2}-2x-2$. Find the points of intersection.

${\begin{aligned}y&=x^{2}-2x-2\\y&=3x-8\\3x-8&=x^{2}-2x-2\\x^{2}-5x+6&=0\\(x-2)(x-3)&=0\\x&=\{2,3\}\\y&=\{-2,1\}\\\therefore {\text{The points of }}&{\text{intersection are at }}(2,-2){\text{ and }}(3,1)\end{aligned}}$

In some cases, we need to find a constant that ensures there is only one point of intersection. In said cases, we should use the discriminant.

e.g. Find the value $k$ such that the line $y=x+k$ is tangent to $y=x^{2}-3x-2$

${\begin{aligned}Tangent\implies one\ &root\implies \Delta =0\\y&=x^{2}-3x-2\\y&=x+k\\x+k&=x^{2}-3x-2\\x^{2}-4x-2-k&=0\\\Delta =(-4)^{2}-4(1)(-2-k)&=0\\16+8+4k&=0\\4k&=-24\\k&=-6\end{aligned}}$