NBT Mathematics: Analytical Geometry Flashcards

Calculate the distance between points A(−1, 3) and B(5, −5).

Answer: 10

Using the distance formula: d = √[(5−(−1))² + (−5−3)²] = √[6² + (−8)²] = √[36 + 64] = √100 = 10.

M is the midpoint of P(−4, 6) and Q(2, −2). What are the coordinates of M?

Answer: (−1, 2)

Midpoint M = ((x₁+x₂)/2, (y₁+y₂)/2) = ((−4+2)/2, (6+(−2))/2) = (−2/2, 4/2) = (−1, 2).

A line has a gradient of 3/4. What is the gradient of a line perpendicular to it?

Answer: −4/3

For perpendicular lines, m₁ × m₂ = −1. So m₂ = −1 ÷ (3/4) = −4/3. The perpendicular gradient is the negative reciprocal of the original.

A circle has centre (3, −2) and radius 5. Which equation represents this circle?

Answer: (x − 3)² + (y + 2)² = 25

The standard form is (x − h)² + (y − k)² = r², where (h, k) is the centre and r is the radius. With centre (3, −2) and r = 5: (x − 3)² + (y − (−2))² = 25 → (x − 3)² + (y + 2)² = 25.

Are the points A(0, 0), B(2, 3) and C(4, 7) collinear?

Answer: No, because the gradients AB and BC are different

Gradient AB = (3−0)/(2−0) = 3/2. Gradient BC = (7−3)/(4−2) = 4/2 = 2. Since 3/2 ≠ 2, the gradients differ, so the three points are NOT collinear — they do not lie on the same straight line.

Find the equation of the straight line passing through (1, 4) and (3, −2).

Answer: y = −3x + 7

Gradient m = (−2 − 4)/(3 − 1) = −6/2 = −3. Using point (1, 4): y − 4 = −3(x − 1) → y = −3x + 3 + 4 → y = −3x + 7.