Coordinate Geometry (2 dimensions only)

Explanation

• The gradient of a straight line measures vertical change per unit horizontal change. For two points $(x_1,y_1)$ and $(x_2,y_2)$, $m=\frac{y_2-y_1}{x_2-x_1}$.
• Both subtractions must use the same point order; reversing both gives the same result. A positive gradient rises from left to right, a negative gradient falls, and a horizontal line has gradient $0$.
• A vertical line has undefined gradient because its horizontal change is zero.
• Gradient may also be read from $y=mx+c$ after an equation is rearranged.
• Examiners expect the coordinate substitution and simplification, not an unsupported value read approximately from a diagram.

Explanation

• Distinct non-vertical parallel lines have equal gradients. Distinct vertical lines are also parallel, although their gradients are undefined.
• For two non-vertical perpendicular lines, the gradients satisfy $m_1m_2=-1$; equivalently, one is the negative reciprocal of the other. Horizontal and vertical lines form the special perpendicular pair.
• To prove that lines are parallel or perpendicular, calculate both gradients exactly and state the comparison: equal gradients establish parallel lines, while a product of $-1$ establishes perpendicular lines.
• A diagram alone is not proof.
• When an equation is not in $y=mx+c$ form, it should first be rearranged so its gradient can be identified reliably.

Explanation

• The horizontal and vertical coordinate differences between two points form the perpendicular sides of a right-angled triangle. Pythagoras’ theorem therefore gives the distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.
• The same point order should be used for both differences, although squaring makes either consistent order equivalent. Negative differences must be squared with brackets.
• If the square root simplifies, an exact integer or surd should be given unless a decimal accuracy is requested.
• Distances can establish equal sides, calculate a perimeter or verify a geometric property.
• Examiners expect the squared differences to be shown; reading a length from a coordinate grid is insufficient unless the scale and exact endpoints make it explicit.

Explanation

• The midpoint of endpoints $(x_1,y_1)$ and $(x_2,y_2)$ is found by averaging corresponding coordinates: $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$. More generally, if $AP:PB=m:n$, point $P$ lies $\frac{m}{m+n}$ of the way from $A$ to $B$.
• The vector method is $P=A+\frac{m}{m+n}(B-A)$, applied to both coordinates.
• Equivalently, the endpoint coordinates receive opposite segment weights.
• The resulting point should lie between the endpoints for an internal division ratio.
• Examiners expect the ratio to be applied separately and consistently to the $x$- and $y$-coordinates, with a midpoint recognised as the special ratio $1:1$.

Explanation

• In the straight-line form $y=mx+c$, $m$ is the gradient and the line crosses the $y$-axis at $(0,c)$. A line of gradient $m$ through $(x_1,y_1)$ can be written in point-gradient form as $y-y_1=m(x-x_1)$.
• When two points are given, calculate the gradient first, then substitute either point.
• Equivalent forms such as $ax+by=c$ describe the same line and may be required by the question.
• Intercepts are found by setting the other coordinate to zero.
• Examiners expect the equation to satisfy the supplied point or points, so substitution provides a useful exact check and exposes sign errors introduced while expanding brackets.

Explanation

• Two distinct points determine a straight line, although a third exact point is a useful check. For $y=mx+c$, plot the $y$-intercept $(0,c)$ and use the gradient as rise over run to locate another point.
• For $ax+by=c$, either rearrange to $y=mx+c$ or substitute convenient $x$-values to build a table.
• Intercepts are often efficient, but they must lie within the displayed axes.
• Plot points accurately, use a ruler and draw one straight line across the full stated interval.
• Examiners award graph marks for a correct straight line in the required range; separate short joins, a freehand curve or an unlabelled line outside the interval can lose accuracy marks.

Explanation

• The equation $x^2+y^2=r^2$ represents the circle with centre $(0,0)$ and positive radius $r$.
• A point lies on the circle exactly when its coordinates satisfy the equation.
• Coordinate problems may combine this equation with circle facts: an angle in a semicircle is $90^\circ$; the perpendicular from the centre to a chord bisects that chord; a radius is perpendicular to the tangent at its endpoint; and tangents from the same external point have equal lengths.
• The radius is $\sqrt{r^2}$, not $r^2$.
• Examiners expect the relevant circle theorem to be named or applied alongside coordinate calculations rather than inferred only from the appearance of a diagram.

Explanation

• The equation $(x-a)^2+(y-b)^2=r^2$ represents a circle with centre $(a,b)$ and positive radius $r$. The bracket signs are opposite to the centre coordinates: $(x+2)^2$ corresponds to centre coordinate $-2$.
• Given a centre and a point on the circle, calculate $r^2$ as the squared distance between them.
• Given diameter endpoints, their midpoint is the centre and one quarter of the squared diameter is $r^2$.
• A candidate point can be checked by substitution.
• Examiners expect the final equation in centre-radius form with both brackets, the correct signs and $r^2$ on the right; expanding the equation is unnecessary unless specifically requested.

Explanation

• A tangent touches a circle at one point and is perpendicular to the radius at that point. For a non-axis-aligned radius, calculate its gradient from the centre to the point of contact, then take the negative reciprocal for the tangent gradient.
• A vertical radius gives a horizontal tangent, while a horizontal radius gives a vertical tangent.
• The tangent equation follows from point-gradient form $y-y_1=m(x-x_1)$ using the point of contact.
• For a circle centred away from the origin, the radius gradient must start at the actual centre.
• Examiners expect the perpendicular-gradient step and the substitution of the contact point; using the radius gradient unchanged produces the equation of the radius, not the tangent.