13.3 Analytic Geometry
Key Takeaways
- Distance is the Pythagorean length between two points; midpoint averages the coordinates.
- Gradient is rise over run; parallel lines share it; perpendicular gradients multiply to −1.
- The equation of BN is a single unscaffolded job: compute the gradient, write point-gradient form, clear fractions, and check the second point.
- A vertical line has undefined gradient and equation x = k; it cannot be written as y = mx + c.
- The circle (x − h)² + (y − k)² = r² is used when a diameter or a centre and radius are implied; compare a point’s left-hand side with r² to decide on, inside, or outside.
13.3 Analytic Geometry
Quick Answer: Analytic geometry on MAT is distance, midpoint, gradient, the equation of a line, parallel and perpendicular lines, and, when needed, the equation of a circle. A typical stem names two points B and N and asks for the equation of BN in one shot: no first find the gradient, hence find the equation. Work from original coordinates by hand; there is no calculator.
Unscaffolded: equation of BN, with new coordinates
NSC analytic-geometry questions are often a ladder: (a) gradient of BN, (b) hence the equation of BN, (c) hence a midpoint or a parallel. MAT Test Content asks you to apply principles of analytic geometry as a single job. Independent OpenExamPrep practice below uses a fresh pair B(4, −3) and N(−2, 5) — not a reprint of any official figure, and not a claim of official review.
Gradient of BN:
m = (5 − (−3)) / (−2 − 4) = 8 / (−6) = −4/3.
Point-gradient form at B:
y − (−3) = (−4/3)(x − 4)
y + 3 = (−4/3)(x − 4)
Multiply by 3: 3y + 9 = −4x + 16
4x + 3y − 7 = 0.
Check N: 4(−2) + 3(5) − 7 = −8 + 15 − 7 = 0. Both given points lie on the line, so the algebra survived. A sign error on the gradient, using +4/3 through B, produces 4x − 3y − 25 = 0, which contains B but misses N: 4(−2) − 3(5) − 25 = −8 − 15 − 25 ≠ 0. Swapping the coefficients 4 and 3 without changing the constant will not pass both points either. The stem never said hence; you still had to find m yourself.
Distance and midpoint first, because the stem may skip naming them
For general A(x₁, y₁) and B(x₂, y₂):
- Distance AB = √[ (x₂ − x₁)² + (y₂ − y₁)² ]
- Midpoint = ( (x₁ + x₂)/2 , (y₁ + y₂)/2 )
Worked example. A(−2, 1), B(4, −3).
Δx = 6, Δy = −4.
AB = √(36 + 16) = √52 = 2√13.
Leaving the answer as √52 is correct but not simplified; options often show 2√13. The squares eat the sign of Δy, so using +4 instead of −4 is harmless for distance. Adding 6 + 4 = 10 is not a distance. √(6 + 4) = √10 forgets to square. √(36 − 16) = √20 subtracts inside the square root.
Midpoint M of AB: ( (−2 + 4)/2 , (1 + (−3))/2 ) = (2/2, (−2)/2) = (1, −1).
The trap (1, 2) averages the x-coordinates but mishandles the y-average (for example 1 − (−3) without dividing, or 1 + 1). The trap (3, −1) uses half the difference of the x-coordinates as if it were the mean, starting from the origin rather than from A. The trap (2, −2) is (Δx/3, Δy/2) or (Δx/2, Δy) without averaging from A. M will reappear as a circle centre when AB is a diameter.
Gradient, parallel, perpendicular
m_AB = (−3 − 1) / (4 − (−2)) = (−4) / 6 = −2/3.
- A line parallel to AB has gradient −2/3.
- A line perpendicular to AB has gradient 3/2, because (−2/3) × (3/2) = −1.
The negative reciprocal of −2/3 is 3/2, not −3/2. Keeping the minus on the reciprocal is the usual perpendicular trap. The reciprocal without the sign change is 2/3, which is neither parallel nor perpendicular.
Horizontal lines have gradient 0. Vertical lines have undefined gradient and equation x = constant. The line through A(−2, 1) and T(−2, 7) is x = −2; you cannot write y = mx + c for it. If a stem asks for the equation of AT, the answer is that vertical line, not a rearranged slope form.
Worked example — parallel through C(6, 5).
y − 5 = (−2/3)(x − 6)
3(y − 5) = −2(x − 6)
3y − 15 = −2x + 12
2x + 3y − 27 = 0.
Check C: 12 + 15 − 27 = 0. A point such as (0, 9) also lies on it: 0 + 27 − 27 = 0. Using +2/3 would not be parallel to AB.
Worked example — perpendicular to AB through N(−2, 5).
y − 5 = (3/2)(x + 2)
2(y − 5) = 3(x + 2)
2y − 10 = 3x + 6
3x − 2y + 16 = 0.
Check N: 3(−2) − 2(5) + 16 = −6 − 10 + 16 = 0. A stem can ask for that perpendicular in one sentence: the equation of the line through N perpendicular to AB, without first asking for m_AB.
Forms of a line
- Point-gradient: y − y₁ = m(x − x₁)
- Two-point: compute m first, then point-gradient
- Standard: ax + by + c = 0
- Slope-intercept: y = mx + k, when the line is not vertical
MAT options may mix these. 4x + 3y − 7 = 0 is the same as y = (−4/3)x + 7/3. Clearing fractions is safer for checking a given point, because substituting into 4x + 3y − 7 is integer arithmetic at B and N.
Intersection, original numbers
BN is 4x + 3y − 7 = 0. A second line is y = 2x − 1.
Substitute: 4x + 3(2x − 1) − 7 = 0
4x + 6x − 3 − 7 = 0
10x = 10, so x = 1, y = 1.
The intersection is (1, 1). Check on BN: 4 + 3 − 7 = 0. Check on the second line: 2(1) − 1 = 1. If substitution produced a contradiction 0 = 5, the lines would be parallel and distinct. If every x worked, they would be the same line. Solving by making y the subject of BN, y = (−4/3)x + 7/3, then setting that equal to 2x − 1, is the same system; the integer route above is faster by hand.
Circle when the stem needs it
The circle with centre (h, k) and radius r:
(x − h)² + (y − k)² = r².
If AB is a diameter, the centre is the midpoint (1, −1) and r² = (half of AB)² = (√13)² = 13, so
(x − 1)² + (y + 1)² = 13.
Point A: (−2 − 1)² + (1 + 1)² = 9 + 4 = 13, on the circle. Point B likewise.
A point P(4, 2): (4 − 1)² + (2 + 1)² = 9 + 9 = 18 > 13, outside. A value less than 13 is inside. Expanding the circle equation gives x² + y² − 2x + 2y − 11 = 0, which is useful if options are expanded. Completing the square in reverse recovers the centre: the −2x term belongs to (x − 1)², the +2y term belongs to (y + 1)².
The angle in a semicircle is a right angle: if P lies on this circle, angle APB = 90°. That is a property you may need without a hence show ladder; it is still analytic geometry, not the Euclidean circle-theorem chapter that follows.
Distance from a point to a line
From (x₀, y₀) to ax + by + c = 0,
distance = |ax₀ + by₀ + c| / √(a² + b²).
From the origin to 4x + 3y − 7 = 0: |−7| / 5 = 7/5.
From C(6, 5) to BN: |24 + 15 − 7| / 5 = 32/5.
The 3-4-5 of the coefficients 3 and 4 is why the denominator is 5. Recognising that triple keeps the arithmetic exact on a calculator-free paper. Dropping the absolute value can produce a negative distance, which is not a length. Using √(16 + 9) = √25 is required; using 4 + 3 = 7 in the denominator is the QL-style add-the-edges error applied to coefficients.
Method
- Sketch the given points so the sign of the gradient is visible (BN falls as x increases: negative gradient).
- Compute m, then write y − y₁ = m(x − x₁), then clear denominators.
- Substitute the second given point as a check.
- Parallel: copy m. Perpendicular: negative reciprocal, including the sign change.
- For a circle, write centre and r² from the stem (diameter, or a centre plus a point).
- Never wait for a hence that will not arrive.
Traps
- Finding m and then stopping, as if the stem had asked only for the gradient
- Writing the perpendicular gradient with the wrong sign
- Averaging x but not y, or reporting (Δx, Δy) as the midpoint
- Expanding (x − 1)² as x² − 1
- Treating a vertical line as y = constant
- Using a calculator to divide 8 by −6 into a rounded decimal when the options are fractions
Independent OpenExamPrep coordinates in this section are A(−2, 1), B(4, −3), N(−2, 5), and C(6, 5). They are teaching numbers, not an NBTP diagram.
The midpoint of A(−2, 1) and B(4, −3) is
The line through A(−2, 1) and B(4, −3) has gradient −2/3. The gradient of a line perpendicular to AB is
The points are B(4, −3) and N(−2, 5). The equation of BN is