Kerala SSLC (Class 10) Linear Equations in One Variable — practice questions — page 2

Explore 37 free Linear Equations in One Variable questions in Kerala SSLC (Class 10) Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Linear Equations in One Variable questions and exercises with answers

Showing questions 31–37 of 37.

CLASS10 · Mathematics · Linear Equations — ConsistencyRead the answer and worked solution →The equation of the straight line joining the points (2, 4) and (5, 6) is:Read the answer and worked solution →The slopes of two perpendicular lines satisfy which relation?Read the answer and worked solution →The equation of a line joining (0, 0) and (1, 1) is:Read the answer and worked solution →The equation of the line which contains every point of the form (x, 2x + 3) is:Read the answer and worked solution →Tap the steps in order to find the equation of the line joining (2, 4) and (5, 6).Read the answer and worked solution →Pick the line whose slope (steepness) is HIGHER.Read the answer and worked solution →

Worked practice questions

CLASS10 · Mathematics · Linear Equations — Consistency

Read exercise and reference answers

Items

Item
Lines intersect at exactly one point
Answer
left
Item
Parallel lines
Answer
right
Item
a₁/a₂ ≠ b₁/b₂
Answer
left
Item
a₁/a₂ = b₁/b₂ ≠ c₁/c₂
Answer
right
Item
Unique x, y satisfy both equations
Answer
left
Item
Graphs never meet
Answer
right

Categories

Left: Consistent (one solution)
Right: Inconsistent (no solution)

Instructions

Classify each scenario: consistent or inconsistent system?

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The equation of the straight line joining the points (2, 4) and (5, 6) is:

2x − 3y + 8 = 0
x − 3y + 6 = 0
3x − 2y + 2 = 0
x + y − 6 = 0

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Correct answer: A. 2x − 3y + 8 = 0

1. Slope between (2, 4) and (5, 6) is (6 − 4)/(5 − 2) = 2/3. 2. Point-slope form using (2, 4): y − 4 = (2/3)(x − 2). 3. Multiply through by 3: 3(y − 4) = 2(x − 2) → 3y − 12 = 2x − 4. 4. Rearrange to 2x − 3y + 8 = 0. 5. The point (11, 10) satisfies 22 − 30 + 8 = 0, confirming the equation. _Source: SCERT Kerala Std X Mathematics Part-2, Chapter 9-11 (pp 227-246, 2019 ed.)._

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The slopes of two perpendicular lines satisfy which relation?

Slopes are equal (m₁ = m₂)
Slopes' product is −1 (m₁ m₂ = −1)
Slopes sum to 1 (m₁ + m₂ = 1)
Slopes are reciprocals (m₁ = 1/m₂)

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Correct answer: B. Slopes' product is −1 (m₁ m₂ = −1)

1. Suppose lines of slopes p and q are perpendicular at point (a, b). 2. Pick (a+1, b+p) on the first line and (a+1, b+q) on the second; these and (a, b) form a right triangle. 3. By Pythagoras: (1² + p²) + (1² + q²) = (q − p)². 4. Simplifying gives 2 + 2pq = 0, hence pq = −1. 5. So the slopes of perpendicular lines multiply to −1. _Source: SCERT Kerala Std X Mathematics Part-2, Chapter 9-11 (pp 227-246, 2019 ed.)._

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