CBSE Class 9 Maths Ganita Manjari Part 2 Ch 13 Two Variables One Line Very Short Answer Questions with Answers

Very Short Answer Questions

Question 1. Write 2.5x − 0.4y = 1.2 in standard form with integer coefficients, and state the values of a, b and c.
Answer:
Multiply by 10: 25x − 4y = 12.
Standard form: 25x − 4y − 12 = 0, so a = 25, b = −4, c = −12.

Teacher's Note:
1. One decimal place → multiply by 10.
2. c takes the sign after moving across.

Question 2. Write x = 9 as a linear equation in two variables in standard form, and state the values of a, b and c.
Answer:
x + 0·y − 9 = 0, so a = 1, b = 0, c = −9.
This is allowed because a and b are not both zero. Its graph is the vertical line through (9, 0).

Teacher's Note:
1. Give y a zero coefficient.
2. Link to 'not both zero'.

Question 3. State, with a reason, whether 5x − \( \frac{3}{y} \) = 2 is a linear equation in two variables.
Answer:
It is not linear.
The term 3/y has y in the denominator (power −1). Multiplying by y gives a 5xy term, where the variables are multiplied.

Teacher's Note:
1. Variables in denominators are not allowed.
2. Clearing the fraction does not help.

Question 4. Find two solutions of 4x − 7y = 28, one with y = 0 and one with x = 0.
Answer:
y = 0: 4x = 28, x = 7 → (7, 0).
x = 0: −7y = 28, y = −4 → (0, −4).

Teacher's Note:
1. These are the intercept points.
2. Watch the negative sign.

Question 5. Determine whether (−3, 2) is a solution of 5x + 4y + 7 = 0, showing your working.
Answer:
5(−3) + 4(2) + 7 = −15 + 8 + 7 = 0.
The left side is 0, so (−3, 2) is a solution.

Teacher's Note:
1. Substitute into the whole left side.
2. Result 0 means the point is on the line.

Question 6. The ordered pair (2, −3) is a solution of kx + 5y = −7. Find the value of k.
Answer:
2k + 5(−3) = −7, so 2k − 15 = −7 and 2k = 8.
k = 4. Check: 8 − 15 = −7 ✓.

Teacher's Note:
1. Substitute, then solve for k.
2. Always check.

Question 7. Find the coordinates of the two points at which the line 6x − 5y = 30 meets the axes.
Answer:
x-axis (y = 0): 6x = 30, x = 5 → (5, 0).
y-axis (x = 0): −5y = 30, y = −6 → (0, −6).

Teacher's Note:
1. Two intercepts are enough to draw the line.
2. Put one variable zero at a time.

Question 8. Explain why 3x + 2y = 12 has no solution in which x and y are both negative.
Answer:
If x and y were both negative, 3x and 2y would both be negative, so 3x + 2y would be negative.
But it must equal 12, which is positive. So no such solution exists.
The line never enters the third quadrant.

Teacher's Note:
1. Sign argument is enough.
2. Link to quadrants on the graph.

Question 9. Find a solution of 2x − 5y = 10 that lies in the fourth quadrant, and state why your point lies in that quadrant.
Answer:
Take y = −1: 2x + 5 = 10, so x = 5/2.
Point (5/2, −1): x > 0 and y < 0, so it is in the fourth quadrant.

Teacher's Note:
1. Choose a negative y and find x.
2. Fourth quadrant: x positive, y negative.

Question 10. Write 7x − 3y + 15 = 0 in the form y = mx + d, and state the values of m and d.
Answer:
3y = 7x + 15, so y = \( \frac{7}{3} \)x + 5.
m = \( \frac{7}{3} \), d = 5.

Teacher's Note:
1. Check with m = −a/b and d = −c/b.
2. Positive slope: line rises.

Question 11. Find the slope of the line through the points (5, −2) and (−1, 7).
Answer:
Slope = \( \frac{7 - (-2)}{-1 - 5} \) = \( \frac{9}{-6} \) = \( -\frac{3}{2} \).

Teacher's Note:
1. Keep the order the same top and bottom.
2. Negative: line falls left to right.

Question 12. The points (1, 4), (3, 10) and (6, 19) lie on one line. Show this by finding the slope from the first two points and then from the last two.
Answer:
(1, 4) to (3, 10): 6/2 = 3.
(3, 10) to (6, 19): 9/3 = 3.
Equal slopes through the shared point (3, 10), so the points are collinear.

Teacher's Note:
1. Mention the common point.
2. Equal slopes alone are not enough without it.

Question 13. The guideline for a wheelchair ramp is a rise of 1 cm for every 12 cm of horizontal length. Find the horizontal length needed for a rise of 21 cm.
Answer:
Slope = 1/12, so run = 12 × rise.
Run = 12 × 21 = 252 cm (2.52 m).

Teacher's Note:
1. Rise/run is fixed.
2. Convert units if asked.

Question 14. State the slope and the y-intercept of the line y = 4, and explain why the line x = 4 cannot be written in the form y = mx + d.
Answer:
y = 4: slope 0, y-intercept 4 (horizontal line).
x = 4 is vertical: its slope is undefined (division by zero), and it does not give one y for each x. So it cannot be written as y = mx + d.

Teacher's Note:
1. Horizontal: slope 0.
2. Vertical: slope undefined.

Question 15. Write the equation of the line of slope −3 whose y-intercept is 7, first in the form y = mx + d and then in standard form.
Answer:
y = −3x + 7.
Standard form: 3x + y − 7 = 0.

Teacher's Note:
1. Check: −a/b = −3 and −c/b = 7.
2. Keep a positive in standard form.

Question 16. The line y = mx + 5 passes through the point (2, 11). Find the value of m.
Answer:
11 = 2m + 5, so 2m = 6 and m = 3.
The line is y = 3x + 5.

Teacher's Note:
1. Substitute the point.
2. Check: 3(2) + 5 = 11.

Question 17. Show that the lines 2x + 3y = 7 and 4x + 6y = 1 have the same slope.
Answer:
2x + 3y = 7 → y = \( -\frac{2}{3} \)x + \( \frac{7}{3} \); slope \( -\frac{2}{3} \).
4x + 6y = 1 → y = \( -\frac{4}{6} \)x + \( \frac{1}{6} \); slope \( -\frac{2}{3} \).
Slopes are equal; intercepts differ, so the lines are parallel.

Teacher's Note:
1. Simplify −4/6.
2. Different intercepts → parallel, not coincident.

Question 18. The sum of two numbers is 37 and their difference is 9. Frame a pair of linear equations for this, taking the numbers as x and y.
Answer:
Taking x as the larger number: x + y = 37 and x − y = 9.

Teacher's Note:
1. State which number is larger.
2. Sum and difference give two equations.

Question 19. A rickshaw fare consists of a fixed charge of ₹x together with ₹y for every kilometre travelled. A 5 km ride costs ₹75 and an 8 km ride costs ₹108. Frame a pair of linear equations in x and y.
Answer:
x + 5y = 75 and x + 8y = 108.
The fixed charge is paid once; y is paid for each kilometre.

Teacher's Note:
1. Fixed charge has coefficient 1.
2. Distance multiplies the per-km rate.

Question 20. Solve by the method of substitution: x + 2y = 11 and 3x − y = 12.
Answer:
From the first, x = 11 − 2y.
3(11 − 2y) − y = 12 → 33 − 7y = 12 → y = 3.
x = 11 − 6 = 5. Solution (5, 3).

Teacher's Note:
1. Isolate the variable with coefficient 1.
2. Check in the second equation.

Question 21. In the pair 2x + 3y = 19 and 5x − 3y = 16, one variable can be eliminated by adding the two equations as they stand. Name that variable, and solve the pair.
Answer:
y is eliminated (coefficients +3 and −3).
Adding: 7x = 35, x = 5. Then 10 + 3y = 19, y = 3.
Solution (5, 3).

Teacher's Note:
1. Opposite coefficients → add.
2. Check: 25 − 9 = 16.

Question 22. In the pair 4x − 3y = 14 and 4x + 5y = −2, one variable can be removed without multiplying either equation by anything. Name that variable, give the reason, and carry out that one elimination step, stopping as soon as a single-variable equation is reached.
Answer:
x can be removed, since its coefficient is 4 in both equations.
Subtracting: (4x − 3y) − (4x + 5y) = 14 − (−2), giving −8y = 16.

Teacher's Note:
1. Equal coefficients → subtract.
2. Stop where the question says.

Question 23. Using the ratios of the coefficients, state the nature of the solutions of 3x − 5y + 8 = 0 and 6x − 10y + 16 = 0.
Answer:
a₁/a₂ = 3/6 = 1/2, b₁/b₂ = −5/−10 = 1/2, c₁/c₂ = 8/16 = 1/2.
All equal: infinitely many solutions (coincident lines).

Teacher's Note:
1. Second equation is twice the first.
2. Write all three ratios.

Question 24. Show that 2x + 7y − 3 = 0 and 6x + 21y − 5 = 0 have no solution, by trying to eliminate y and stating what the elimination step leaves behind.
Answer:
Multiply the first by 3: 6x + 21y − 9 = 0.
Subtract the second: −9 − (−5) = 0, i.e. −4 = 0.
This is false, so the pair has no solution (parallel lines).

Teacher's Note:
1. Both variables vanish; the leftover decides.
2. False statement → no solution.

Question 25. A pair of linear equations in two variables is known to have a unique solution. State what this tells you about the ratios \( \frac{a_1}{a_2} \) and \( \frac{b_1}{b_2} \), and state what it tells you about \( \frac{c_1}{c_2} \).
Answer:
It tells us \( \frac{a_1}{a_2} \) ≠ \( \frac{b_1}{b_2} \).
It tells us nothing about \( \frac{c_1}{c_2} \); any constants still give one crossing point.

Teacher's Note:
1. Only the first two ratios decide 'unique'.
2. Example: x + 2y + 1 = 0, x + 3y + 1 = 0.

Question 26. Find the value of k for which 2x + ky = 5 and 6x + 9y = 15 represent coincident lines.
Answer:
a₁/a₂ = 2/6 = 1/3 and c₁/c₂ = 5/15 = 1/3.
So k/9 = 1/3, giving k = 3. Check: 3(2x + 3y = 5) is 6x + 9y = 15 ✓.

Teacher's Note:
1. All three ratios must match.
2. Verify by multiplying.

Question 27. Find the value of k that must be avoided if kx − 2y = 3 and 4x + 3y = 7 are to meet at exactly one point.
Answer:
Need k/4 ≠ −2/3.
k/4 = −2/3 gives 3k = −8, so avoid k = \( -\frac{8}{3} \).

Teacher's Note:
1. Set the ratios equal to find the bad value.
2. At that value slopes are equal.

Question 28. The lines ℓ₁ and ℓ₂ in the figure below have been drawn with the same steepness.
State how many solutions the corresponding pair of equations has, and give the reason.
Answer:
No solution.
Same steepness means equal slopes; they cross the y-axis at different heights, so the gap between them is the same everywhere and never closes. No point lies on both lines.

Teacher's Note:
1. Equal slope + different intercept = parallel.
2. Parallel lines have no common point.

Question 29. Write down a pair of linear equations in two variables whose only solution is (3, −2), and show that your pair is satisfied by that ordered pair.
Answer:
For example, x + y = 1 and x − y = 5.
3 + (−2) = 1 ✓ and 3 − (−2) = 5 ✓.
Since 1/1 ≠ 1/(−1), the solution is unique.

Teacher's Note:
1. Many correct pairs exist.
2. Make sure the ratios a₁/a₂ and b₁/b₂ differ.

Question 30. Explain why a pair of linear equations in two variables can never have exactly two solutions.
Answer:
Each equation is a straight line; a solution is a common point.
If two lines shared two points, both would be the one line through those points, so they would coincide and share infinitely many points.
So the only possibilities are 0, 1 or infinitely many.

Teacher's Note:
1. Two points fix one line.
2. List the three possible outcomes.