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Gradients, midpoints, partitions and areas used as proofs, and the equations of circles and parabolas.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson you put figures on a pair of axes and prove things about them by computing rather than by reasoning from congruent triangles: equal gradients for parallel sides, gradients multiplying to $-1$ for perpendicular ones, averages for midpoints, and a base and a height for an area. You then write the equation of a circle from its centre and radius, recover a hidden centre and radius by completing the square, and meet the parabola as the set of points equidistant from a focus and a directrix.
You can find the gradient of a line through two points, you know that parallel lines have equal gradients, and you can find a distance with $a^2 + b^2 = c^2$. Putting a figure on axes turns every geometric claim about it into one of those three calculations.
Coordinate proof: proving a general fact by placing the figure on axes and computing, rather than by reasoning from congruent triangles.
Gradient (slope): the change in $y$ divided by the change in $x$.
Perpendicular: at right angles. Two gradients are perpendicular when their product is $-1$.
Midpoint: the average of the two endpoints, coordinate by coordinate.
Partition in the ratio $1 : 2$: the point one third of the way along, because the two pieces are one part and two parts of three.
Base and height: the side you measure along, and the distance to the opposite vertex measured at right angles to it.
Put a figure on axes and its properties become arithmetic. Parallel lines have equal gradients, so showing both pairs of opposite sides have equal gradients proves a quadrilateral is a parallelogram. Perpendicular gradients multiply to $-1$, so a line of gradient $\frac{2}{3}$ is crossed at right angles by one of gradient $-\frac{3}{2}$. The midpoint averages the coordinates, and the point dividing a segment in the ratio $1 : 2$ sits one third of the way along: from $(1, 2)$ to $(7, 11)$ the steps are $6$ and $9$, a third of which are $2$ and $3$, giving $(3, 5)$. An area comes from a base and a height: the triangle with vertices $(0, 0)$, $(6, 0)$ and $(2, 5)$ has base $6$ along the axis and height $5$, so its area is $15$. Choosing where to put the figure — a corner at the origin, a side along an axis — is part of the proof, because it decides how much arithmetic you have to do.
Another way: picture
A parallelogram on a grid with a corner at the origin. The step from one corner to the next is drawn as a right-angled staircase of $2$ across and $3$ up, and the identical staircase is drawn again on the opposite side.
Another way: story
A coordinate proof is a geometric claim translated into arithmetic. Once the figure is on the grid you never look at it again; the numbers carry the argument.
Turning the gradient over but forgetting the sign. The perpendicular of $\frac{2}{3}$ is $-\frac{3}{2}$, not $\frac{3}{2}$. Multiply your answer by the original: it has to come out at exactly $-1$.
Reading a ratio of $1 : 2$ as a half. It splits the segment into $1 + 2 = 3$ equal pieces, so the point is one third of the way along. Add the two parts of the ratio before you divide anything.
Using a slanted side as the height. The height of a triangle is measured at right angles to the base you chose. Choosing the side that lies along an axis makes the height simply the third vertex's distance from that axis, which is why placing the figure carefully is half of a coordinate proof.
$A = (1, 1)$, $B = (4, 2)$, $C = (6, 6)$. The step from $B$ to $C$ is $2$ across and $4$ up.
Opposite sides are parallel and equal, so that step repeats from $A$.
From $A = (1, 1)$, that step lands on $D = (3, 5)$.
Check: the step from $A$ to $B$ is $3$ across, $1$ up, and from $D$ to $C$ it is also $3$ across, $1$ up.
Both pairs of opposite sides match.
Two vertices are on the $x$-axis, so take that side as the base: $6$.
Choose the base that lies flat.
The height is the third vertex's distance from the axis: $5$.
Area $= \frac{1}{2} \times 6 \times 5 = 15$.
Turn it over: $\frac{5}{4}$.
Change the sign: $-\frac{5}{4}$, and $\frac{4}{5} \times -\frac{5}{4} = -1$.
$ABCD$ is a parallelogram with $A = (-4, -4)$, $B = (-3, -6)$ and $C = (-2, -9)$. Plot $D$.
Plot your answer on the grid:
Find the point that divides the segment from $(5, 2)$ to $(8, 5)$ in the ratio $1 : 2$, measured from the first point.
(x, y)
A line has gradient $3/8$. What gradient is perpendicular to it?
answer
A triangle has vertices $(0, 0)$, $(4, 0)$ and $(2, 3)$. What is its area?
answer
You can complete the square on a quadratic, and you know that the distance between two points comes from $a^2 + b^2 = c^2$. A circle's equation is nothing but that distance formula with the distance held fixed, and completing the square is how you read a centre out of an equation that hides it.
Centre-radius form: $(x - h)^2 + (y - k)^2 = r^2$, the equation of the circle of radius $r$ about $(h, k)$.
General form: the same circle multiplied out, $x^2 + y^2 + Dx + Ey + F = 0$, with the centre hidden.
Completing the square: rewriting $x^2 + Dx$ as $(x + \frac{D}{2})^2 - (\frac{D}{2})^2$, which is how the centre is recovered.
Focus and directrix: a point and a line. A parabola is the set of points the same distance from both.
Locus: the set of all points obeying a condition — which is what an equation of a curve really says.
A circle is the set of points a fixed distance from a fixed point, so the distance formula gives its equation directly: centre $(3, -2)$ and radius $5$ is $(x - 3)^2 + (y + 2)^2 = 25$. Read backwards, the numbers subtracted inside the brackets are the centre and the number on the right is the radius squared. When the equation arrives multiplied out — $x^2 + y^2 - 6x + 4y - 3 = 0$ — complete the square in each variable: $x^2 - 6x = (x - 3)^2 - 9$ and $y^2 + 4y = (y + 2)^2 - 4$, so the equation becomes $(x - 3)^2 + (y + 2)^2 = 16$, a circle of centre $(3, -2)$ and radius $4$. To test whether a point lies on a circle, substitute it: $(5, 1)$ in $(x - 2)^2 + (y + 3)^2 = 25$ gives $9 + 16 = 25$, so it does. A parabola is the set of points equidistant from a focus and a directrix; $y = \frac{x^2}{4p}$ has focus $(0, p)$ and directrix $y = -p$, so $y = \frac{x^2}{4}$ has its focus at $(0, 1)$.
Another way: picture
A circle drawn on axes with its centre marked at $(3, -2)$ and one radius drawn to a point on the edge. The radius is the hypotenuse of a right triangle whose legs are $x - 3$ and $y + 2$.
Another way: steps
From centre and radius: subtract each coordinate inside a bracket, square each bracket, and square the radius on the right. From an equation: halve each linear coefficient, change its sign for the centre, and add back what completing the square removed.
Reading the signs inside the brackets as the centre. $(x - 3)^2 + (y + 2)^2 = 25$ has centre $(3, -2)$, not $(3, 2)$ or $(-3, 2)$. The form subtracts the centre, so a plus sign inside means a negative coordinate.
Giving $r^2$ as the radius. The right-hand side of the equation is the radius squared: $25$ means a radius of $5$. And going the other way, a radius of $5$ must be written as $25$.
Forgetting to put back what completing the square took out. Turning $x^2 - 6x$ into $(x - 3)^2$ adds $9$ that was not there, so $9$ has to come off the other side. Losing it moves the radius without moving the centre, and the answer looks plausible.
The form subtracts the centre: $(x - 3)^2 + (y - (-2))^2$, which is $(x - 3)^2 + (y + 2)^2$.
A negative coordinate becomes a plus sign.
The right-hand side is $5^2 = 25$.
$x^2 - 6x = (x - 3)^2 - 9$: half of $-6$ is $-3$.
Completing the square.
$y^2 + 4y = (y + 2)^2 - 4$.
$(x - 3)^2 + (y + 2)^2 - 9 - 4 - 3 = 0$, so $(x - 3)^2 + (y + 2)^2 = 16$: centre $(3, -2)$, radius $4$.
The $16$ is $r^2$, so $r = 4$.
Subtract the centre inside the brackets: $(x + 1)^2 + (y - 4)^2$.
Square the radius on the right: $= 9$.
Write the equation of the circle with centre $(-2, -4)$ and radius $8$.
(x - h)^2 + (y - k)^2 = s
The circle $x^2 + y^2 + 0x + 0y - 64 = 0$ has centre $(h, k)$ and radius $r$. Fill them in.
centre (h, k), radius r
The parabola $y = \frac{x^2}{4}$ has its focus at $(0, p)$ and its directrix at $y = -p$. What is $p$?
Answer:
Write the equation of the circle with centre $(7, -5)$ and radius $4$.
(x - h)^2 + (y - k)^2 = s
On a map where one unit is one kilometre, a phone mast stands at $(-2, -3)$ and reaches $9$ km in every direction. A village sits at $(6, 3)$. Is the village covered?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
$ABCD$ is a parallelogram with $A = (-4, -1)$, $B = (-2, 3)$ and $C = (-1, 1)$. Plot $D$.
Plot your answer on the grid:
The circle $x^2 + y^2 + 6x - 8y - 24 = 0$ has centre $(h, k)$ and radius $r$. Fill them in.
centre (h, k), radius r
You can prove properties of figures with coordinates and write and read the equations of circles and parabolas. Explain why the number on the right of a circle's equation is the radius squared, and why a ratio of $1 : 2$ puts the point one third of the way along.
8. Your turn: the gradient perpendicular to $\frac{4}{5}$, step 2
19. Your turn: centre $(-1, 4)$, radius $3$, step 2