Analytical Geometry: Distance, Midpoint and Gradient — Worksheet & Memo

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Grade 10 · Analytical Geometry

Analytical Geometry: Distance, Midpoint and Gradient

MEMORANDUM

Name:

Class:

Date:

Mark

/ 29

Answer all questions. Show all your working — marks are awarded for method as well as the final answer. Teacher copy: accept any correct equivalent method.

1
Distance between (1 ; 1) and (4 ; 5). (distance)
(2)
√(9 + 16) = √25 = 5✓✓ (2)
2
Distance between (−2 ; 1) and (1 ; 5). (distance)
(2)
√(9 + 16) = 5✓✓ (2)
3
Midpoint of (2 ; 4) and (6 ; 10). (midpoint)
(2)
(2 + 6)/2,(4 + 10)/2 = (4 ; 7)✓✓ (2)
4
Gradient of the line through (1 ; 2) and (5 ; 10). (gradient)
(2)
(10 − 2)/(5 − 1) = 2✓✓ (2)
5
Are gradients 3 and 3 parallel or perpendicular? (parallel)
(1)
Equal → parallel✓ (1)
6
What gradient is perpendicular to gradient 2? (perpendicular)
(1)
−1/2 (negative reciprocal)✓ (1)
7
What gradient is perpendicular to gradient −3/4? (perpendicular)
(1)
4/3✓ (1)
8
Are (0 ; 0), (2 ; 4), (3 ; 6) collinear? (collinear)
(3)
Gradients 2 and 2 → Yes✓✓✓ (3)
9
Midpoint of (1 ; 2) and (x ; 8) is (4 ; 5). Find x. (unknown)
(2)
(1 + x)/2 = 4 → x = 7✓✓ (2)
10
Distance from (0 ; 0) to (x ; 12) is 13 (x > 0). Find x. (unknown)
(3)
x² + 144 = 169 → x = 5✓✓✓ (3)
11
P(−1 ; 1), Q(3 ; 3), R(5 ; 7) and S(1 ; 5) are the vertices of a quadrilateral. Prove that PQRS is a parallelogram.
(4)
m(PQ) = 2/4 = ½ ✓; m(SR) = 2/4 = ½ ✓, so PQ ∥ SR; PQ = √20 and SR = √20 ✓; ∴ parallelogram (one pair opp sides equal and parallel) ✓(4)
12
A(0 ; 0), B(4 ; 2) and C(2 ; 6). Show that AB ⊥ BC.
(3)
m(AB) = 2/4 = ½ ✓; m(BC) = (6 − 2)/(2 − 4) = −2 ✓; ½ × (−2) = −1, so AB ⊥ BC ✓(3)
13
The diagonals of quadrilateral KLMN are KM and LN. K(1 ; 1), L(6 ; 2), M(7 ; 6), N(2 ; 5). Show that the diagonals bisect each other.
(3)
Midpoint KM = (4 ; 3,5) ✓; midpoint LN = (4 ; 3,5) ✓; the midpoints are the same, so the diagonals bisect each other ✓(3)

TOTAL: 29 marks