Class 9 · Maths · Linear Equations in Two Variables

Equations of Lines Parallel to the Axes MCQs

Practice Equations of Lines Parallel to the Axes multiple-choice questions from Linear Equations in Two Variables (Class 9 Maths) - tap an answer for instant feedback and a step-by-step solution. Practice the full set free on the RankByte app.

Equations of Lines Parallel to the AxesQuiz - Solve & Score

  1. Q1. The four lines x = 1, x = 6, y = 2, y = 5 bound a rectangle. The enclosed region measures how many square units?

    • A.15
    • B.16
    • C.8
    • D.20

    Answer: A. 15

    Quick parse - a typical math numerical. The data on the table: 1, 6, 2, 5. We are after how many square units?. Tool of choice - width = |6-1| = 5. Rearrange it for the unknown before substituting. Numbers in: width = |6-1| = 5 → height = |5-2| = 3 → area = 5×.3 = 15. Lock in option A) 15. Trap-watch: option B) '16' misses the point - That is the PERIMETER, not the area; option C) '8' is incorrect: Area is width×height, not width+height.

  2. Q2. The four lines x = -2, x = 3, y = 1, y = 7 bound a rectangle. The enclosed region measures how many square units?

    • A.30
    • B.22
    • C.11
    • D.35

    Answer: A. 30

    Given: -2; 3; 1; 7. The unknown asked is: how many square units?. Relation we use - width = |3--2| = 5. This is the equation that links the given quantities to the unknown (math, chapter 'Linear Equations in Two Variables'). Substituting: width = |3--2| = 5 → height = |7-1| = 6 → area = 5×.6 = 30. Hence the answer is A) 30. Wrong options at a glance: option B) '22' is wrong because That is the PERIMETER, not the area; option C) '11' is incorrect: Area is width×height, not width+height; option D) '35' is wrong because Off by one row of width 5; area = 5×6.

  3. Q3. The four lines x = 0, x = 7, y = -1, y = 3 bound a rectangle. The enclosed region measures how many square units?

    • A.28
    • B.22
    • C.11
    • D.35

    Answer: A. 28

    First, read what's there - a typical math numerical. The data on the table: 0, 7, -1, 3. We are after how many square units?. Tool of choice - width = |7-0| = 7. Rearrange it for the unknown before substituting. Numbers in: width = |7-0| = 7 → height = |3--1| = 4 → area = 7×.4 = 28. Lock in option A) 28. Trap-watch: option B) '22' is wrong because That is the PERIMETER, not the area; option C) '11' doesn't hold - Area is width×height, not width+height.

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