NBT Mathematics: Analytical Geometry

Question 1

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

Accepted Answer

10

Answer

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

Question 2

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

Accepted Answer

(−1, 2)

Answer

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

Question 3

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

Accepted Answer

−4/3

Answer

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

Question 4

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

Accepted Answer

(x − 3)² + (y + 2)² = 25

Answer

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.

Question 5

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

Accepted Answer

No, because the gradients AB and BC are different

Answer

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.

(NBT) National Benchmark Test Practice Test

(NBT) National Benchmark Test Practice Test

NBT Mathematics: Analytical Geometry