13.1 Geometry
Key Takeaways
- The distance between two points (x1,y1) and (x2,y2) is sqrt((x2-x1)^2 + (y2-y1)^2), and the midpoint is ((x1+x2)/2, (y1+y2)/2).
- The slope of a line perpendicular to one with slope m is -1/m, so a line perpendicular to 2x - 4y + 11 = 0 (slope 1/2) has slope -2.
- A circle of radius r has circumference 2*pi*r and area pi*r^2; a right triangle with legs a, b and hypotenuse c satisfies a^2 + b^2 = c^2 (Pythagoras).
- Two lines are perpendicular if and only if the product of their slopes is -1 (m1*m2 = -1), and parallel lines have equal slopes.
- The general line equation Ax + By + C = 0 has slope -A/B, which is the fastest route to extracting slope from an implicit form.
13.1 Geometry for the PAF GD Pilot Initial Test
Quick Answer: Geometry questions in the PAF GD Pilot Initial Test are FSc-level problems on lines, triangles, circles, and coordinate geometry. The highest-yield facts are the distance formula, midpoint formula, slope of a line (including the perpendicular-slope rule m1m2 = -1), Pythagoras' theorem, and the circle formulas (circumference 2pir, area pir^2). Roughly 4-6% of the full paper sits here, all inside the 13% Mathematics domain.
Lines and Angles
A line in the plane can be written in three useful forms:
- Slope-intercept: y = mx + c, where m is the slope and c is the y-intercept.
- Point-slope: y - y1 = m(x - x1), handy when you know one point and the slope.
- General (implicit): Ax + By + C = 0, with slope m = -A/B.
Two lines are parallel when their slopes are equal (m1 = m2) and perpendicular when the product of their slopes is -1 (m1*m2 = -1). A horizontal line has slope 0; a vertical line has undefined slope.
Worked example (perpendicular slope). Find the slope of a line perpendicular to 2x - 4y + 11 = 0.
- Rewrite in slope-intercept form: 4y = 2x + 11, so y = (1/2)x + 11/4. The original slope is m1 = 1/2.
- The perpendicular slope satisfies m1*m2 = -1, so (1/2)*m2 = -1, giving m2 = -2.
This single rule (negative reciprocal) is the most frequently tested geometry fact on the PAF Initial Test.
Triangles
A triangle has three sides and three angles that sum to 180 degrees. Key results:
- Pythagoras' theorem: In a right triangle with legs a, b and hypotenuse c, a^2 + b^2 = c^2.
- Area: (1/2)baseheight. Heron's formula, area = sqrt(s(s-a)(s-b)(s-c)) with s = (a+b+c)/2, works for any triangle when the height is unknown.
- Similar triangles: corresponding angles equal, corresponding sides in proportion (a1/a2 = b1/b2 = c1/c2).
- Congruent triangles: identical in shape and size, proved by SSS, SAS, ASA, or RHS.
Circles
A circle is the set of points at a fixed distance (the radius r) from a centre. The diameter d = 2r, a chord joins two points on the circle, and a tangent touches the circle at exactly one point and is perpendicular to the radius at that point.
- Circumference: C = 2pir = pi*d.
- Area: A = pi*r^2.
- Arc length (central angle theta in degrees): L = (theta/360)2pi*r.
- Sector area: A_sector = (theta/360)pir^2.
Coordinate Geometry
Given two points P(x1, y1) and Q(x2, y2):
- Distance: d = sqrt((x2 - x1)^2 + (y2 - y1)^2).
- Midpoint: M = ((x1 + x2)/2, (y1 + y2)/2).
- Slope of PQ: m = (y2 - y1)/(x2 - x1).
These three formulas are the backbone of almost every coordinate-geometry question on the test. When the line is given in implicit form Ax + By + C = 0, recover the slope immediately as m = -A/B before doing anything else.
Formula Table
| Shape | Quantity | Formula |
|---|---|---|
| Right triangle | Hypotenuse | c = sqrt(a^2 + b^2) |
| Triangle | Area | (1/2)baseheight |
| Triangle | Area (Heron) | sqrt(s(s-a)(s-b)(s-c)), s = (a+b+c)/2 |
| Circle | Circumference | 2pir |
| Circle | Area | pi*r^2 |
| Circle | Arc length (theta deg) | (theta/360)2pi*r |
| Line (slope-intercept) | Slope, y-intercept | m, c in y = mx + c |
| Line (general) | Slope | -A/B in Ax + By + C = 0 |
| Two lines | Perpendicular condition | m1*m2 = -1 |
| Two points | Distance | sqrt((x2-x1)^2 + (y2-y1)^2) |
| Two points | Midpoint | ((x1+x2)/2, (y1+y2)/2) |
Circle Equation (Coordinate-Geometry Extension)
The standard circle with centre (h, k) and radius r is (x - h)^2 + (y - k)^2 = r^2. The expanded form x^2 + y^2 + Dx + Ey + F = 0 represents a circle provided D^2 + E^2 - 4F > 0, with centre (-D/2, -E/2) and radius (1/2)*sqrt(D^2 + E^2 - 4F). Completing the square recovers the centre and radius, and this technique is routinely tested.
Worked example. Find the centre and radius of x^2 + y^2 - 6x + 8y - 11 = 0.
- Group x and y terms: (x^2 - 6x) + (y^2 + 8y) = 11.
- Complete each square: (x^2 - 6x + 9) + (y^2 + 8y + 16) = 11 + 9 + 16 = 36, i.e. (x - 3)^2 + (y + 4)^2 = 36.
- Read off centre (3, -4) and radius r = sqrt(36) = 6.
Note the sign flip: the constant inside (x - 3)^2 gives centre x = +3, and (y + 4)^2 = (y - (-4))^2 gives centre y = -4. A frequent exam trap is to write the centre as (3, 4) by dropping the sign on the y-term.
Common Traps
- Distance formula sign: sqrt((x2 - x1)^2 + (y2 - y1)^2) uses squared differences, so the order of subtraction does not matter. Do not write sqrt((x2 - x1) + (y2 - y1)) — the squares are essential; without them you get a meaningless sum, not a distance.
- Perpendicular vs parallel: a line perpendicular to slope m has slope -1/m, not 1/m. If m = 1/2, the perpendicular is -2, not +2. Candidates under time pressure pick +2 (a common distractor) because they forget the negative reciprocal.
- Area of a circle vs circumference: pir^2 (area) is quadratic in r, 2pi*r (circumference) is linear. With r = 7 and pi = 22/7, area = 154 and circumference = 44 — the exam distractor swaps these two values.
- Heron's semi-perimeter: s = (a + b + c)/2 is half the perimeter, not the full perimeter. Using the full perimeter produces a value 2x too large and breaks the formula.
What is the slope of a line perpendicular to 2x - 4y + 11 = 0?
What is the distance between the points A(1, 2) and B(4, 6)?
A circle has radius 7 cm. What is its area (use pi = 22/7)?
The midpoint of a segment with endpoints P(-3, 4) and Q(5, -2) is: