Coordinate geometry: lines, areas and circles

Three situations where a straight line or a circle appears on a map. The pictures come first; the lines, points and circle drawn over them come from the same numbers as the solutions.

$y = \dfrac{4}{5}x + \dfrac{1}{5}$$A(1,\ 1)$$J(6,\ 5)$$T(9,\ \frac{1}{2})$$y = -\dfrac{3}{2}x + 14$$x + 16y - 17 = 0$

Maths In Real Life 1

A trail and its side trail

Maps of paths and roads are drawn on grids, so a straight trail can be described by one equation, and where two paths meet can be found by calculation instead of measuring.

On a plan of a nature reserve, each grid square stands for ten metres. A straight trail starts at $A(1, 1)$ and passes through $J(6, 5)$. A road runs along the line $x + 16y - 17 = 0$, passing through $A$. A side trail leaves the trail at $J$ and heads towards the road in the direction with gradient $- \frac{3}{2}$ on the plan.

• (a)Find the equation of the trail in the form $y = mx + c$.
• (b)Find the angle that the trail makes with the $x$-axis of the plan (its angle of inclination), correct to $1$ decimal place.
• (c)Find the equation of the side trail in the form $y = mx + c$.
• (d)Find the coordinates of the point $T$ where the side trail meets the road.

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(a)

$\displaystyle m\;= \dfrac{5 - 1}{6 - 1} = \textcolor{#010104}{\frac{4}{5}}$

Gradient = change in $y$ divided by change in $x$ (rise over run).

$\displaystyle y - 1 = \frac{4}{5}\left(x - 1\right)$

Point-slope form: $y - y_1 = m(x - x_1)$.

$\displaystyle y = \frac{4}{5}x - \frac{4}{5} + 1$

Expand the bracket.

$\displaystyle y = \textcolor{#010106}{\frac{4}{5}x + \frac{1}{5}}$

Collect the constants.

(b)

Since $\tan\theta = m$, $\theta = \tan^{-1}(\frac{4}{5}) = 38.7^\circ$ (1 d.p.).

The gradient is positive, so the line leans up to the right and $\theta$ is acute.

(c)

The line goes through the point above and has the given gradient.

$\displaystyle y - 5 = (- \frac{3}{2})\left(x - 6\right)$

Point-slope form: $y - y_1 = m(x - x_1)$.

$\displaystyle y = -\frac{3}{2}x + 9 + 5$

Expand the bracket.

$\displaystyle y = \textcolor{#010106}{-\frac{3}{2}x + 14}$

Collect the constants.

(d)

$\displaystyle 3x + 2y - 28 = 0 \iff y = -\frac{3}{2}x + 14$

Make $y$ the subject of the first equation.

$\displaystyle x + 16y - 17 = 0 \iff y = -\frac{1}{16}x + \frac{17}{16}$

Make $y$ the subject of the second equation.

At the point where the lines meet, both equations give the same $y$, so set the two right-hand sides equal.

$\displaystyle -\frac{3}{2}x + 14 = -\frac{1}{16}x + \frac{17}{16}$

Equal $y$ values: one equation in $x$ only.

$\displaystyle -\frac{3}{2}x + 14 = -\frac{1}{16}x + \frac{17}{16}$

Collect the $x$ terms on the left and the numbers on the right, then divide.

$\displaystyle -\frac{23}{16}x = - \frac{207}{16}$

$\displaystyle x = 9$

$\displaystyle y\;= (- \frac{3}{2}) \times 9 + 14 = \textcolor{#010105}{\frac{1}{2}}$

Put $x$ back into either equation to find $y$.

The lines meet at $(9,\ \frac{1}{2})$.

Answer

(a) $y = \frac{4}{5}x + \frac{1}{5}$

(b) $\theta = 38.7^\circ$ (1 d.p.)

(c) $y = -\frac{3}{2}x + 14$

(d) $(9,\ \frac{1}{2})$

$5$$10$$15$$20$$2$$4$$6$$8$$10$$O$$x$$y$$y = 4x - 31$$A(1,\ 7)$$B(17,\ 3)$$P(7,\ -3)$$y = -3$

Maths In Real Life 2

A bus stop the same distance from two exits

A new bus stop is often placed so that walkers from two nearby exits have the same distance to cover. Every point equally far from two places lies on one straight line: the perpendicular bisector of the line joining them.

On a plan, two MRT exits are at $A(1, 7)$ and $B(17, 3)$. A main road runs along the line $y = -3$. A new bus stop $P$ on the road is to be exactly the same distance from $A$ and from $B$.

• (a)Express the perpendicular bisector of $AB$ in the form $y = mx + c$.
• (b)Find the coordinates of $P$.

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(a)

The perpendicular bisector of $AB$ passes through the midpoint of $AB$ and is perpendicular to $AB$.

$\displaystyle x_M\;= \dfrac{1 + 17}{2} = \dfrac{18}{2} = \textcolor{#010105}{9}$

Midpoint: the average of the two $x$-coordinates.

$\displaystyle y_M\;= \dfrac{7 + 3}{2} = \dfrac{10}{2} = \textcolor{#010105}{5}$

And the average of the two $y$-coordinates.

So the midpoint is $M = (9, 5)$.

$\displaystyle m_1\;= \dfrac{3 - 7}{17 - 1} = \dfrac{-4}{16} = \textcolor{#010104}{- \frac{1}{4}}$

Gradient = change in $y$ divided by change in $x$ (rise over run).

$\displaystyle m_2\;= \textcolor{#010104}{4}$

Gradient of the bisector: negative reciprocal of the gradient of $AB$.

$\displaystyle y - 5 = 4\left(x - 9\right)$

Point-slope form through the midpoint.

$\displaystyle y = 4x - 36 + 5$

Expand the bracket.

$\displaystyle y = \textcolor{#010106}{4x - 31}$

Collect the constants.

(b)

$\displaystyle 4x - y - 31 = 0 \iff y = 4x - 31$

Make $y$ the subject of the first equation.

$\displaystyle y + 3 = 0 \iff y = -3$

Make $y$ the subject of the second equation.

At the point where the lines meet, both equations give the same $y$, so set the two right-hand sides equal.

$\displaystyle 4x - 31 = -3$

Equal $y$ values: one equation in $x$ only.

$\displaystyle 4x - 31 = -3$

Collect the $x$ terms on the left and the numbers on the right, then divide.

$\displaystyle 4x = 28$

$\displaystyle x = 7$

$\displaystyle y\;= 4 \times 7 - 31 = \textcolor{#010105}{-3}$

Put $x$ back into either equation to find $y$.

The lines meet at $(7,\ -3)$.

Answer

(a) $y = 4x - 31$

(b) $(7,\ -3)$

$5$$4$$O$$x$$y$$y = \dfrac{5}{9}x - 4$$C(2,\ 3)$$P(0,\ -4)$$Q(9,\ 1)$$r = \sqrt{53}$

Maths In Real Life 3

How much of a path has Wi-Fi?

In a simple model, the signal of a hotspot reaches the same distance in every direction, so its coverage is drawn as a circle. A straight path through that circle is a chord, and its length tells you how much of the path is covered.

On a plan of a park, each grid square stands for ten metres. A Wi-Fi hotspot is at $C(2, 3)$ and its signal reaches a distance of $\sqrt{ 53 }$ units in every direction. A straight path has equation $y = \frac{5}{9}x - 4$.

• (a)Write down the equation of the circle that marks the edge of the signal, in the form $(x - a)^2 + (y - b)^2 = r^2$.
• (b)The path enters the covered region at $P$ and leaves it at $Q$. Find the coordinates of $P$ and $Q$, and the length of the path inside the covered region, in units and in metres (to the nearest metre).

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(a)

A circle with centre $(a,\ b)$ and radius $r$ has equation $(x - a)^2 + (y - b)^2 = r^2$.

$\displaystyle r^2\;= \left(\sqrt{53}\right)^2 = \textcolor{#010105}{53}$

Square the radius.

$\displaystyle (x - 2)^2 + (y - 3)^2 = 53$

Put the centre and $r^2$ into $(x - a)^2 + (y - b)^2 = r^2$.

(b)

Where the line meets the circle, the point satisfies both equations. Replace $y$ in the circle equation by the expression for $y$ from the line, to get an equation in $x$ only.

$\displaystyle (x - 2)^2 + \left(\frac{5}{9}x - 4 - 3\right)^2 = 53$

Put $mx + c$ in place of $y$.

$\displaystyle (x - 2)^2 + \left(\frac{5}{9}x - 4 - 3\right)^2 = 53 \iff x^{2} - 9 x = 0$

Expand both brackets, take $r^2$ across, and divide by any common factor.

$\displaystyle x^{2} - 9 x = 0$

Factorise and use the zero-product rule.

$\displaystyle x \left(x - 9\right) = 0$

$\displaystyle x = 0 \text{ or } x = 9$

$\displaystyle y\;= \frac{5}{9} \times 0 - 4 = \textcolor{#010105}{-4}$

Put $x = 0$ into the equation of the line.

$\displaystyle y\;= \frac{5}{9} \times 9 - 4 = \textcolor{#010105}{1}$

Put $x = 9$ into the equation of the line.

The line meets the circle at $(0,\ -4)$ and $(9,\ 1)$.

$\displaystyle PQ\;= \sqrt{\left(9 - 0\right)^2 + \left(1 - (-4)\right)^2} = \sqrt{81 + 25} = \sqrt{106}$

Distance formula between the two points of the chord.

Answer

(a) $(x - 2)^2 + (y - 3)^2 = 53$

(b) $(0,\ -4)$ and $(9,\ 1)$; chord length $= \sqrt{106}$; In metres: $\sqrt{ 106 } \times 10 \approx 103$ m.

Coordinate geometry puts a picture into numbers. A point is a pair of coordinates, a line is an equation, and questions about parallel lines, right angles and distances become arithmetic that you can check. Everything starts with the gradient, which says how steeply a line climbs.

Two lines are parallel when they climb at the same rate, so parallel lines have equal gradients. Two lines are perpendicular when they meet at a right angle, and then the gradients are linked by a surprisingly neat rule.

Three points are collinear when they lie on one straight line. Because the gradient is the same everywhere on a line, three points $A$, $B$, $C$ are collinear exactly when the gradient of $AB$ equals the gradient of $BC$ (the two segments share the point $B$). This test needs the segments not to be vertical: points with the same $x$-coordinate are collinear on the line $x = c$.

The midpoint of a segment is the point exactly halfway along it. Its coordinates are the averages of the end-point coordinates. The same idea of shared halves explains a useful fact about parallelograms: their diagonals cut each other in half.

Sometimes a point splits a segment in a given ratio, say $AR : RB = m : n$. Draw a right-angled triangle on $AB$: the smaller triangle on $AR$ is similar to it, so the run and the rise are both shared in the ratio $m : (m + n)$. The same idea of similar triangles can extend a line to meet an axis.

To find the equation of a straight line you need a gradient and a point. When two points are given, the gradient comes from them. When the line is described as parallel or perpendicular to another, the gradient comes from that line. In every case, finish with the point-slope form $y - y_1 = m(x - x_1)$ and tidy it up.