Coordinate Geometry — Practice

Question 1

Find the distance between (2, 3) and (5, 7).

Accepted Answer

Step 1 — √[(5−2)² + (7−3)²] = √(9 + 16) = √25.

✦ Answer: 5 units.

Question 2

Find the midpoint of the segment joining (−2, 4) and (6, −8).

Accepted Answer

Step 1 — ((−2+6)/2, (4−8)/2) = (2, −2).

✦ Answer: (2, −2).

Question 3

Are (1, 1), (2, 3) and (3, 5) collinear?

Accepted Answer

Step 1 — Area = ½|1(3−5) + 2(5−1) + 3(1−3)| = ½|−2 + 8 − 6| = 0.

✦ Answer: Yes, collinear (area = 0).

Question 4

Find the point dividing the join of (−1, 7) and (4, −3) in the ratio 2 : 3.

Accepted Answer

Step 1 — x = (2·4 + 3·(−1))/5 = (8−3)/5 = 1.

Step 2 — y = (2·(−3) + 3·7)/5 = (−6+21)/5 = 3.

✦ Answer: (1, 3).

Question 5

Find the area of the triangle with vertices (0, 0), (4, 0) and (0, 3).

Accepted Answer

Step 1 — ½|0(0−3) + 4(3−0) + 0(0−0)| = ½|12| = 6.

✦ Answer: 6 square units.

Question 6

Find x if the distance between (x, 2) and (3, 6) is 5 units.

Accepted Answer

Step 1 — (x−3)² + (6−2)² = 25 ⇒ (x−3)² + 16 = 25.

Step 2 — (x−3)² = 9 ⇒ x − 3 = ±3 ⇒ x = 6 or x = 0.

✦ Answer: x = 6 or x = 0.

Question 7

In what ratio does the x-axis divide the segment joining (2, −3) and (5, 6)?

Accepted Answer

Step 1 — On the x-axis, y = 0. Let ratio k:1. y = (6k + (−3))/(k+1) = 0.

Step 2 — 6k − 3 = 0 ⇒ k = 1/2, so ratio = 1 : 2.

✦ Answer: 1 : 2.

Question 8

Show that A(1, 7), B(4, 2), C(−1, −1), D(−4, 4) are the vertices of a square.

Accepted Answer

Step 1 — AB = √(9+25) = √34; BC = √(25+9) = √34; CD = √(9+25) = √34; DA = √(25+9) = √34 (all sides equal).

Step 2 — Diagonals: AC = √(4+64) = √68; BD = √(64+4) = √68 (equal).

Step 3 — Equal sides AND equal diagonals ⇒ square.

✦ Answer: it is a square (side √34, diagonal √68).

Question 9

Two vertices of a triangle are (3, −5) and (−7, 4). If its centroid is (2, −1), find the third vertex.

Accepted Answer

Step 1 — Centroid x: (3 + (−7) + x₃)/3 = 2 ⇒ −4 + x₃ = 6 ⇒ x₃ = 10.

Step 2 — Centroid y: (−5 + 4 + y₃)/3 = −1 ⇒ −1 + y₃ = −3 ⇒ y₃ = −2.

✦ Answer: (10, −2).

Question 10

Find k if the points (7, −2), (5, 1) and (3, k) are collinear.

Accepted Answer

Step 1 — Area = 0: 7(1−k) + 5(k−(−2)) + 3(−2−1) = 0.

Step 2 — 7 − 7k + 5k + 10 − 9 = 0 ⇒ −2k + 8 = 0 ⇒ k = 4.

✦ Answer: k = 4.

Practice — Coordinate Geometry

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