Chapter Exam

Chapter 7: Coordinate Geometry — Online MCQ Test

Q: In the ordered pair (7, -3), the abscissa is:

Answer: A. 7 — The abscissa is the x-coordinate of a point. In (7, -3), the x-coordinate is 7.

Q: The distance between two points (x1, y1) and (x2, y2) is given by:

Answer: A. √[(x2 - x1)^2 + (y2 - y1)^2] — The distance formula is derived from Pythagoras' theorem. It is √[(x2 - x1)^2 + (y2 - y1)^2].

Q: Find the midpoint of the line segment joining (-2, 5) and (4, -1).

Answer: A. (1, 2) — Midpoint = ((-2 + 4)/2, (5 + (-1))/2) = (1, 2).

Q: In what ratio does the point P(2, 3) divide the line segment joining A(1, 2) and B(4, 5)?

Answer: A. 1:2 — Let AP:PB = m:n. Using the x-coordinate, (4m + n)/(m + n) = 2, which gives n = 2m. Hence, the ratio is 1:2.

Q: A point P(x, y) is equidistant from A(2, 3) and B(6, 7). Which equation must P satisfy?

Answer: A. x + y = 9 — Equating distances gives (x - 2)^2 + (y - 3)^2 = (x - 6)^2 + (y - 7)^2. Simplifying gives x + y = 9.

MATHS · CLASS 10 · Andhra State Board

Practice Chapter 7: Coordinate Geometry with a free chapter-wise online MCQ test for Andhra State Board CLASS 10 MATHS. This chapter covers: Cartesian plane Coordinates Distance formula Section formula Midpoint formula Area of triangle Collinear points Ordered pairs Graph plotting. AI-generated questions from basic to board-exam level, with instant results and explanations.

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Chapter 7: Coordinate Geometry — Important Questions & Answers (FAQ)

Frequently asked questions from Andhra State Board CLASS 10 MATHS — Chapter 7: Coordinate Geometry, with answers and explanations. These are sample questions; the exam has its own separate question set.

In the ordered pair (7, -3), the abscissa is:

A. 7 ✓
B. -3
C. 10
D. -10

Answer: A. 7
The abscissa is the x-coordinate of a point. In (7, -3), the x-coordinate is 7.

The distance between two points (x1, y1) and (x2, y2) is given by:

A. √[(x2 - x1)^2 + (y2 - y1)^2] ✓
B. (x2 - x1) + (y2 - y1)
C. √[(x2 + x1)^2 + (y2 + y1)^2]
D. [(x2 - x1)^2 - (y2 - y1)^2]

Answer: A. √[(x2 - x1)^2 + (y2 - y1)^2]
The distance formula is derived from Pythagoras' theorem. It is √[(x2 - x1)^2 + (y2 - y1)^2].

Find the midpoint of the line segment joining (-2, 5) and (4, -1).

A. (1, 2) ✓
B. (2, 1)
C. (-1, 3)
D. (3, -1)

Answer: A. (1, 2)
Midpoint = ((-2 + 4)/2, (5 + (-1))/2) = (1, 2).

In what ratio does the point P(2, 3) divide the line segment joining A(1, 2) and B(4, 5)?

A. 1:2 ✓
B. 2:1
C. 1:3
D. 3:1

Answer: A. 1:2
Let AP:PB = m:n. Using the x-coordinate, (4m + n)/(m + n) = 2, which gives n = 2m. Hence, the ratio is 1:2.

A point P(x, y) is equidistant from A(2, 3) and B(6, 7). Which equation must P satisfy?

A. x + y = 9 ✓
B. x - y = 9
C. x + y = 5
D. 2x + y = 9

Answer: A. x + y = 9
Equating distances gives (x - 2)^2 + (y - 3)^2 = (x - 6)^2 + (y - 7)^2. Simplifying gives x + y = 9.

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