Coordinate Geometry

Three things you can always find from two points

Give a question two points, $A(x_1,y_1)$ and $B(x_2,y_2)$, and it can ask for three quantities before it asks for anything else: the distance $AB$, the gradient of $AB$, and the midpoint of $AB$. All three come from the same two numbers, the run $x_2-x_1$ and the rise $y_2-y_1$, so write those down first. Almost every lost mark in this topic starts with a sign error inside one of them.

The distance is Pythagoras applied to the right-angled triangle whose legs are the run and the rise, so $AB=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$. The gradient is rise over run, $m=\frac{y_2-y_1}{x_2-x_1}$. The midpoint is the average of the two ends, $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$. Squaring in the distance formula removes the signs, so the order of the two points does not matter there. It matters everywhere else.

For $A(-1,2)$ and $B(7,8)$: the run is $7-(-1)=8$ and the rise is $8-2=6$. The distance is $\sqrt{64+36}=\sqrt{100}=10$, the gradient is $\frac{6}{8}=\frac{3}{4}$, and the midpoint is $(\frac{-1+7}{2},\frac{2+8}{2})=(3,5)$. Check the midpoint by eye: it should sit halfway between the two points, and $(3,5)$ does.

The three forms of the equation of a straight line

The syllabus names three forms and expects you to move between them. The slope-intercept form $y=mx+c$ is the one to finish in when a question asks for the equation of a line: $m$ is the gradient and $c$ is the y-intercept, the value of $y$ where the line crosses the y-axis.

The point-gradient form $y-y_1=m(x-x_1)$ is the one to start in, because it needs exactly the two things a question usually gives you: a gradient and one point on the line. Nearly every line question is two lines of working in this form.

The general form $ax+by+c=0$ is the tidiest way to write an answer containing fractions. Multiply through by the denominators so that $a$, $b$ and $c$ are integers. The form is part of the question: a correct line written in the wrong form loses the final mark, and it is one of the easiest marks on the paper to throw away.

• $y=\frac{3}{4}x+\frac{5}{2}$ is the same line as $3x-4y+10=0$: multiply by 4 and rearrange
• $y-3=-2(x+1)$ is the same line as $y=-2x+1$: expand and tidy up
• $x=4$ is a vertical line and cannot be written as $y=mx+c$

Finding the equation of a line from sufficient information

Sufficient information means enough to pin the line down. That is two points, or one point and the gradient. From two points you find the gradient first and then use either point in $y-y_1=m(x-x_1)$. You never need both points for the second step, and using the one with the smaller numbers saves arithmetic.

For the line through $A(2,3)$ and $B(6,6)$: the gradient is $\frac{6-3}{6-2}=\frac{3}{4}$, so $y-3=\frac{3}{4}(x-2)$, which tidies to $y=\frac{3}{4}x+\frac{3}{2}$, or $3x-4y+6=0$ in general form. Check it against the point you did not use: at $x=6$, $\frac{3}{4}\times 6+\frac{3}{2}=4.5+1.5=6$, which is the y-coordinate of $B$.

The two conditions that are not a gradient behave the same way once you have written the gradient down. 'Parallel to $y=4x-7$' hands you $m=4$. 'Perpendicular to $AB$' hands you the negative reciprocal of the gradient of $AB$. From there it is substitution, exactly as before.

• Two points: find the gradient from the run and the rise, then substitute one point
• One point and the gradient: straight into $y-y_1=m(x-x_1)$
• One point and a parallel line: copy the gradient of the given line
• One point and a perpendicular line: use $m_2=-\frac{1}{m_1}$, then substitute

Parallel and perpendicular lines

Parallel lines have equal gradients: $m_1=m_2$. They are still different lines, so their intercepts differ. A line parallel to $y=2x+3$ passing through $(0,-1)$ is $y=2x-1$: same gradient, different intercept.

Perpendicular lines have gradients whose product is $-1$, so $m_2=-\frac{1}{m_1}$. Flip the gradient over and change its sign: perpendicular to $\frac{3}{4}$ is $-\frac{4}{3}$, and perpendicular to $-2$ is $\frac{1}{2}$. Negating alone is not enough, and it is the commonest error in the topic. A vertical line is perpendicular to a horizontal one, and that case has to be handled by thinking rather than by the formula.

The perpendicular bisector of $AB$ is the line through the midpoint of $AB$ at right angles to $AB$. It is also the locus of the points that are the same distance from $A$ as from $B$, so a question about a point 'equidistant from two given points' is a perpendicular bisector question in disguise. For $A(-1,2)$ and $B(7,8)$: the midpoint is $(3,5)$, the gradient of $AB$ is $\frac{3}{4}$, so the bisector has gradient $-\frac{4}{3}$ and equation $y-5=-\frac{4}{3}(x-3)$, that is $4x+3y-27=0$.

Where two graphs meet

A point of intersection satisfies both equations at once, so finding one means solving a pair of simultaneous equations. Two straight lines meet in one point unless they are parallel. A line and a curve meet in up to two points. The number of solutions is the number of intersections, and a question can ask you for that number without asking for the points themselves.

For the lines $2x+3y=12$ and $y=x-1$, substitute the second into the first: $2x+3(x-1)=12$, so $5x-3=12$, $x=3$ and then $y=2$. The intersection is $(3,2)$. When one equation is already written as $y=\dots$, substituting is faster than eliminating, and it is the method the mark scheme expects.

A line and a quadratic curve end in a quadratic equation, and the discriminant $b^2-4ac$ then counts the intersections before you find them: two intersections when $b^2-4ac>0$, one when it is zero, and none when it is negative. The one-solution case is a tangent, and 'find the values of $k$ for which the line touches the curve' is always a discriminant-equals-zero question.

Here is the whole routine on a circle. The line $y=x+k$ meets the circle $x^2+y^2=8$ where $x^2+(x+k{)}^2=8$, that is $2x^2+2kx+k^2-8=0$. Its discriminant is $4k^2-8(k^2-8)=64-4k^2$. That is zero at $k=\pm 4$, positive for $-4<k<4$ and negative outside. So the line is a tangent for $k=\pm 4$, cuts the circle twice between those values, and misses it entirely beyond them.

• Line and line: one intersection, or none if the gradients are equal
• Line and curve: substitute, rearrange to $a{x}^2+bx+c=0$, then use $b^2-4ac$
• $b^2-4ac>0$: two intersections, so the line cuts the curve
• $b^2-4ac=0$: one intersection, so the line is a tangent
• $b^2-4ac<0$: no intersection, so the line misses the curve

The circle: centre, radius and the expanded form

A circle is the set of points a fixed distance from a fixed point, so its equation comes straight from the distance formula. The points a distance $r$ from $(a,b)$ satisfy $\sqrt{(x-a{)}^2+(y-b{)}^2}=r$, which squares to $(x-a{)}^2+(y-b{)}^2=r^2$. Read it off: the centre is $(a,b)$ and the radius is $r$, not $r^2$. The sign flip inside the bracket is the classic slip - $(x-3{)}^2$ means the x-coordinate of the centre is 3, not $-3$.

Cambridge also uses the expanded form $x^2+y^2+2gx+2fy+c=0$. To get the centre and radius back, complete the square in $x$ and in $y$. For $x^2+y^2-6x+8y-11=0$: $x^2-6x=(x-3{)}^2-9$ and $y^2+8y=(y+4{)}^2-16$, so $(x-3{)}^2+(y+4{)}^2=9+16+11=36$. The centre is $(3,-4)$ and the radius is 6.

In general, $x^2+y^2+2gx+2fy+c=0$ is the circle with centre $(-g,-f)$ and radius $\sqrt{g^2+f^2-c}$. That formula is worth knowing for a quick check, but completing the square is the safer route in an exam: the method marks are for the completing the square, and it leaves a page the examiner can follow.

Tangents, chords and the angle in a semicircle

Line-and-circle problems are solved with three geometrical facts, all of which you already know. A tangent touches the circle at exactly one point and is perpendicular to the radius drawn to that point. The perpendicular from the centre to a chord bisects the chord. And the angle in a semicircle is a right angle, so if $AB$ is a diameter then every point $P$ on the circle gives $\angle APB={90}^{\circ }$.

The tangent fact turns into algebra most often. Find the gradient of the radius from the centre to the point of contact, take the negative reciprocal, and you have the gradient of the tangent. For the circle $x^2+y^2=5$ at the point $P(2,1)$: the radius $OP$ has gradient $\frac{1}{2}$, so the tangent has gradient $-2$ and equation $y-1=-2(x-2)$, that is $y=-2x+5$.

The perpendicular-from-the-centre fact gives you the midpoint of a chord, and with the radius it gives the distance from the centre to the chord by Pythagoras. The angle in a semicircle is the bridge between the two halves of the topic: if a circle has diameter $AB$, its centre is the midpoint of $AB$ and its radius is half of $AB$, so a coordinate question about a circle can start with nothing more than two points.

The circle on a diameter, and circles through given points

If $AB$ is a diameter of a circle then the centre is the midpoint of $AB$ and the radius is half the distance $AB$. Nothing else is needed: midpoint, then half-length, then write the equation down. For $A(-1,2)$ and $B(7,8)$ the centre is $(3,5)$, $AB=10$ so $r=5$, and the circle is $(x-3{)}^2+(y-5{)}^2=25$.

That same circle is the locus of the points $P$ for which $\angle APB={90}^{\circ }$, which is the angle in a semicircle used backwards. If a question tells you an angle is a right angle and asks where the vertex can be, the answer is a circle with the two other points as the ends of a diameter.

When a circle is given in expanded form and a point is offered as lying on it, substitute. The point lies on the circle exactly when the left-hand side comes to zero. For $x^2+y^2-4x+6y-12=0$ and the point $(5,1)$: $25+1-20+6-12=0$, so $(5,1)$ is on the circle. In a 'show that' question that substitution, written out, is the whole answer - and there is no answer box to fill in, because the working is the answer.

• Centre of a circle with diameter $AB$: the midpoint of $AB$
• Radius of a circle with diameter $AB$: $\frac{1}{2}AB$
• A point lies on a circle when its coordinates make the equation true
• An angle of ${90}^{\circ }$ at $P$ means $P$ lies on the circle with diameter $AB$