Class IX Mathematics worksheet on linear equations in two variables with problems and solutions.

Let's solve each question from the given worksheet step by step, with clear explanations.

---

SECTION A: (1 MARK)

#### 1. Find $ m $, if point $ (7, -3) $ lies on the equation $ y - \frac{3}{2} = m(x - \frac{1}{2}) $.

We are given a point $ (7, -3) $ that lies on the line:
$$
y - \frac{3}{2} = m\left(x - \frac{1}{2}\right)
$$

Substitute $ x = 7 $, $ y = -3 $ into the equation:

$$
-3 - \frac{3}{2} = m\left(7 - \frac{1}{2}\right)
$$

$$
-\frac{6}{2} - \frac{3}{2} = m\left(\frac{14}{2} - \frac{1}{2}\right)
$$

$$
-\frac{9}{2} = m \cdot \frac{13}{2}
$$

Now solve for $ m $:

$$
m = \frac{-\frac{9}{2}}{\frac{13}{2}} = -\frac{9}{13}
$$

But the answer provided is $ m = -\frac{24}{47} $. This suggests a typo in the problem or the expected answer. Let's double-check.

Wait — let's recheck the original equation. Is it possible the equation was meant to be:

$$
y - \frac{3}{2} = m(x - \frac{1}{2})
$$

Yes, and we substituted correctly.

But $ -\frac{9}{13} \neq -\frac{24}{47} $. So either the point or the equation is different.

Wait — perhaps there’s a misprint in the question. But assuming the given point $ (7, -3) $ and equation are correct, our calculation is:

$$
m = -\frac{9}{13}
$$

But the expected answer is $ m = -\frac{24}{47} $. That doesn't match.

Let’s suppose the equation is actually:
$$
y - \frac{3}{2} = m(x - \frac{1}{2})
$$
and point $ (7, -3) $

Then:
$$
-3 - 1.5 = m(7 - 0.5) \Rightarrow -4.5 = m(6.5)
\Rightarrow m = -\frac{4.5}{6.5} = -\frac{9}{13}
$$

Still $ -\frac{9}{13} $. So unless the point or equation is different, the given answer $ -\frac{24}{47} $ is incorrect.

But since the problem says "NCERT Exemplar", maybe the original has different values.

Alternatively, perhaps the equation is:
$$
y - \frac{3}{2} = m(x - \frac{1}{2})
$$
and point $ (7, -3) $, then yes, $ m = -\frac{9}{13} $

So unless there’s a typo, the answer should be $ m = -\frac{9}{13} $, not $ -\frac{24}{47} $

But let's assume the answer key says $ -\frac{24}{47} $, so maybe the point is different? Or equation?

Wait — maybe the equation is written as:
$$
y - \frac{3}{2} = m(x - \frac{1}{2})
$$
but the point is not $ (7, -3) $? No, it is.

Alternatively, maybe the equation is:
$$
y + \frac{3}{2} = m(x - \frac{1}{2})?
$$
No, it's $ y - \frac{3}{2} $

I think there's an error in the provided answer. The correct value of $ m $ is $ -\frac{9}{13} $

But since the answer is marked as $ -\frac{24}{47} $, let's skip this for now and proceed.

---

#### 2. Find the value of $ a $ in the equation $ ax + y = 5 $ if $ x = 2 $ and $ y = 3 $

Given: $ ax + y = 5 $, and $ x = 2 $, $ y = 3 $

Substitute:
$$
a(2) + 3 = 5 \Rightarrow 2a = 2 \Rightarrow a = 1
$$

✔ Answer: $ a = 1 $

---

#### 3. If $ x - 4 = \sqrt{3} y $ is written in standard form $ ax + by + c = 0 $, then find $ a, b, c $

Start with:
$$
x - 4 = \sqrt{3} y
$$

Bring all terms to one side:
$$
x - \sqrt{3} y - 4 = 0
$$

Compare with $ ax + by + c = 0 $

So:
- $ a = 1 $
- $ b = -\sqrt{3} $
- $ c = -4 $

✔ Answer: $ a = 1 $, $ b = -\sqrt{3} $, $ c = -4 $

---

SECTION B: (2 MARKS)

#### 4. Represent an equation of a straight line which is parallel to x-axis and at a distance of 2.5 units below it.

A line parallel to the x-axis is horizontal, so its equation is $ y = k $

Since it's 2.5 units below the x-axis, $ y = -2.5 $

So the equation is:
$$
y = -2.5 \quad \text{or} \quad y + 2.5 = 0
$$

Or in standard form: $ 0x + 1y + 2.5 = 0 $

✔ Equation: $ y = -2.5 $

---

#### 5. For the first km, fare is Rs 15 and for successive distance it is Rs 8 per km. Taking distance covered as $ x $ km and total fare as $ y $ (Rs), represent a linear equation in two variables.

Let:
- $ x $ = total distance in km
- $ y $ = total fare in Rs

First km: Rs 15
Remaining distance: $ (x - 1) $ km at Rs 8 per km

So:
$$
y = 15 + 8(x - 1) = 15 + 8x - 8 = 8x + 7
$$

So the linear equation is:
$$
y = 8x + 7 \quad \text{or} \quad 8x - y + 7 = 0
$$

✔ Answer: $ y = 8x + 7 $ or $ 8x - y + 7 = 0 $

---

#### 6. If $ (2, 3) $ and $ (4, 0) $ lie on the graph of $ ax + by + 1 = 0 $, find $ a $ and $ b $

We have:
$$
ax + by + 1 = 0
$$

Plug in $ (2, 3) $:
$$
2a + 3b + 1 = 0 \quad \text{(1)}
$$

Plug in $ (4, 0) $:
$$
4a + 0b + 1 = 0 \Rightarrow 4a = -1 \Rightarrow a = -\frac{1}{4}
$$

Now plug into (1):
$$
2(-\frac{1}{4}) + 3b + 1 = 0 \Rightarrow -\frac{1}{2} + 3b + 1 = 0 \Rightarrow 3b + \frac{1}{2} = 0 \Rightarrow 3b = -\frac{1}{2} \Rightarrow b = -\frac{1}{6}
$$

But the answer says $ a = \frac{1}{4}, b = \frac{1}{6} $ — but signs are wrong.

Wait — check again.

From $ 4a + 1 = 0 \Rightarrow a = -\frac{1}{4} $

Then $ 2(-\frac{1}{4}) + 3b + 1 = 0 \Rightarrow -0.5 + 3b + 1 = 0 \Rightarrow 3b = -0.5 \Rightarrow b = -\frac{1}{6} $

So $ a = -\frac{1}{4}, b = -\frac{1}{6} $

But the expected answer is $ a = \frac{1}{4}, b = \frac{1}{6} $ — probably they wrote the equation as $ ax + by = -1 $, but here it's $ ax + by + 1 = 0 $

So unless the equation is $ ax + by = 1 $, then signs would change.

But as written: $ ax + by + 1 = 0 $

So $ a = -\frac{1}{4}, b = -\frac{1}{6} $

But the answer says $ a = \frac{1}{4}, b = \frac{1}{6} $ — contradiction.

Wait — let's test the answer.

Suppose $ a = \frac{1}{4}, b = \frac{1}{6} $, then equation is:
$$
\frac{1}{4}x + \frac{1}{6}y + 1 = 0
$$

At $ (2, 3) $:
$$
\frac{1}{4}(2) + \frac{1}{6}(3) + 1 = 0.5 + 0.5 + 1 = 2 \ne 0
$$

Not satisfied.

Try $ a = -\frac{1}{4}, b = -\frac{1}{6} $:
$$
-\frac{1}{4}(2) - \frac{1}{6}(3) + 1 = -0.5 - 0.5 + 1 = 0 ✔
$$

At $ (4, 0) $:
$$
-\frac{1}{4}(4) + 0 + 1 = -1 + 1 = 0 ✔
$$

So correct values are:
$$
a = -\frac{1}{4}, \quad b = -\frac{1}{6}
$$

But the given answer is $ a = \frac{1}{4}, b = \frac{1}{6} $ — likely a sign error in the answer key.

But wait — could the equation be $ ax + by = 1 $? Then:
$ 2a + 3b = 1 $, $ 4a = 1 \Rightarrow a = \frac{1}{4} $, then $ 2(\frac{1}{4}) + 3b = 1 \Rightarrow 0.5 + 3b = 1 \Rightarrow 3b = 0.5 \Rightarrow b = \frac{1}{6} $

Ah! So if the equation were $ ax + by = 1 $, then $ a = \frac{1}{4}, b = \frac{1}{6} $

But the given equation is $ ax + by + 1 = 0 $ → $ ax + by = -1 $

So unless it's $ ax + by = 1 $, the answer is wrong.

But the problem says $ ax + by + 1 = 0 $, so correct values are $ a = -\frac{1}{4}, b = -\frac{1}{6} $

But the answer key says $ a = \frac{1}{4}, b = \frac{1}{6} $ — possibly typo.

Let’s move on.

---

#### 7. Find the co-ordinates of the points where the graph of $ 7x - 3y = 4 $ cuts x-axis and y-axis

To find x-intercept: set $ y = 0 $
$$
7x = 4 \Rightarrow x = \frac{4}{7}
$$
So point: $ \left( \frac{4}{7}, 0 \right) $

To find y-intercept: set $ x = 0 $
$$
-3y = 4 \Rightarrow y = -\frac{4}{3}
$$
So point: $ \left( 0, -\frac{4}{3} \right) $

✔ Answer: X-axis: $ \left( \frac{4}{7}, 0 \right) $, Y-axis: $ \left( 0, -\frac{4}{3} \right) $

But the answer given is:
> X axis $ (\frac{4}{7}, 0) $, Y axis $ (0, -\frac{4}{3}) $

Wait — the image shows $ (0, -\frac{4}{3}) $, but the text says $ (0, -\frac{4}{3}) $ — yes.

But in the image, it says $ (0, -\frac{4}{3}) $, but in your text you wrote $ (0, -\frac{4}{3}) $ — okay.

But in the original, it says $ (0, -\frac{4}{3}) $ — correct.

✔ Correct.

---

SECTION C: (3 MARKS)

#### 8. Solve $ \frac{3x+2}{7} + \frac{4x+1}{5} = \frac{2(2x+1)}{3} $

Find LCM of denominators: 7, 5, 3 → LCM = 105

Multiply both sides by 105:
$$
105 \left( \frac{3x+2}{7} + \frac{4x+1}{5} \right) = 105 \left( \frac{2(2x+1)}{3} \right)
$$

$$
15(3x+2) + 21(4x+1) = 35 \cdot 2(2x+1)
$$

Compute:
$$
15(3x+2) = 45x + 30 \\
21(4x+1) = 84x + 21 \\
\text{LHS: } 45x + 30 + 84x + 21 = 129x + 51
$$

RHS:
$$
35 \cdot 2(2x+1) = 70(2x+1) = 140x + 70
$$

So:
$$
129x + 51 = 140x + 70
\Rightarrow 51 - 70 = 140x - 129x
\Rightarrow -19 = 11x
\Rightarrow x = -\frac{19}{11}
$$

But the answer says $ x = 4 $

Check: Plug $ x = 4 $ into original equation:

LHS:
$$
\frac{3(4)+2}{7} + \frac{4(4)+1}{5} = \frac{12+2}{7} + \frac{16+1}{5} = \frac{14}{7} + \frac{17}{5} = 2 + 3.4 = 5.4
$$

RHS:
$$
\frac{2(8+1)}{3} = \frac{2(9)}{3} = \frac{18}{3} = 6
$$

5.4 ≠ 6 → Not valid.

So either the equation is different or the answer is wrong.

Wait — let's recheck the equation:

Given:
$$
\frac{3x+2}{7} + \frac{4x+1}{5} = \frac{2(2x+1)}{3}
$$

But maybe it's:
$$
\frac{3x+2}{7} + \frac{4x+1}{5} = \frac{2(2x+1)}{3}
$$

We did it correctly.

But answer says $ x = 4 $

Try $ x = 4 $: LHS = $ \frac{14}{7} + \frac{17}{5} = 2 + 3.4 = 5.4 $, RHS = $ \frac{2(9)}{3} = 6 $ → no.

Try $ x = 1 $: LHS = $ \frac{5}{7} + \frac{5}{5} = 0.714 + 1 = 1.714 $, RHS = $ \frac{2(3)}{3} = 2 $ → no.

Try $ x = 0 $: LHS = $ \frac{2}{7} + \frac{1}{5} ≈ 0.286 + 0.2 = 0.486 $, RHS = $ \frac{2(1)}{3} ≈ 0.666 $

No.

So likely the equation is different.

Maybe the equation is:
$$
\frac{3x+2}{7} + \frac{4x+1}{5} = \frac{2(2x+1)}{3}
$$

But perhaps it's supposed to be:
$$
\frac{3x+2}{7} + \frac{4x+1}{5} = \frac{2(2x+1)}{3}
$$

Wait — maybe it's:
$$
\frac{3x+2}{7} + \frac{4x+1}{5} = \frac{2(2x+1)}{3}
$$

But we got $ x = -\frac{19}{11} $, not 4.

So either the answer is wrong or the equation is different.

Alternatively, maybe the equation is:
$$
\frac{3x+2}{7} + \frac{4x+1}{5} = \frac{2(2x+1)}{3}
$$

But no.

Perhaps it's:
$$
\frac{3x+2}{7} + \frac{4x+1}{5} = \frac{2(2x+1)}{3}
$$

Same thing.

Unless the right-hand side is $ \frac{2(2x+1)}{3} $, but maybe it's $ \frac{2(2x+1)}{3} $

Wait — perhaps the equation is:
$$
\frac{3x+2}{7} + \frac{4x+1}{5} = \frac{2(2x+1)}{3}
$$

But let's try solving again.

We had:
$$
15(3x+2) + 21(4x+1) = 35 \cdot 2(2x+1)
\Rightarrow 45x + 30 + 84x + 21 = 70(2x+1)
\Rightarrow 129x + 51 = 140x + 70
\Rightarrow -11x = 19 \Rightarrow x = -\frac{19}{11}
$$

So unless the equation is different, answer is wrong.

But the answer says $ x = 4 $, so maybe the equation is:
$$
\frac{3x+2}{7} + \frac{4x+1}{5} = \frac{2(2x+1)}{3}
$$

No.

Perhaps it's:
$$
\frac{3x+2}{7} + \frac{4x+1}{5} = \frac{2(2x+1)}{3}
$$

Same.

Maybe it's a typo and the equation is:
$$
\frac{3x+2}{7} + \frac{4x+1}{5} = \frac{2(2x+1)}{3}
$$

But still.

Alternatively, maybe the right side is $ \frac{2(2x+1)}{3} $, but no.

I think the answer key is wrong.

But let's assume the equation is correct, and our solution is $ x = -\frac{19}{11} $

But since the answer says $ x = 4 $, perhaps the equation is:
$$
\frac{3x+2}{7} + \frac{4x+1}{5} = \frac{2(2x+1)}{3}
$$

Try $ x = 4 $: LHS = $ \frac{14}{7} + \frac{17}{5} = 2 + 3.4 = 5.4 $, RHS = $ \frac{2(9)}{3} = 6 $ → not equal.

Try $ x = 5 $: LHS = $ \frac{17}{7} + \frac{21}{5} ≈ 2.428 + 4.2 = 6.628 $, RHS = $ \frac{2(11)}{3} = \frac{22}{3} ≈ 7.33 $ → no.

Try $ x = 3 $: LHS = $ \frac{11}{7} + \frac{13}{5} ≈ 1.57 + 2.6 = 4.17 $, RHS = $ \frac{2(7)}{3} = \frac{14}{3} ≈ 4.67 $

No.

So I think the answer key is wrong.

But let's skip and go to next.

---

#### 9. Draw the graph of $ y = x $ and $ y = -x $ on the same Cartesian plane. What do you observe?

Graph of $ y = x $: passes through origin, slope 1, goes up-right.

Graph of $ y = -x $: passes through origin, slope -1, goes down-right.

They intersect at origin $ (0, 0) $, and are perpendicular to each other (since product of slopes = $ 1 \times (-1) = -1 $)

✔ Observation: Both lines pass through origin and are perpendicular to each other.

---

#### 10. Draw the graph of $ 2x + 3y = 12 $ and check whether the points $ (4.5, 1) $ and $ (1.5, 3) $ lie on the graph

First, find intercepts:

- x-intercept: $ y = 0 $ → $ 2x = 12 $ → $ x = 6 $
- y-intercept: $ x = 0 $ → $ 3y = 12 $ → $ y = 4 $

Plot points $ (6, 0) $ and $ (0, 4) $, draw line.

Now check:

Point $ (4.5, 1) $:
$$
2(4.5) + 3(1) = 9 + 3 = 12 ✔ \text{ lies on graph}
$$

Point $ (1.5, 3) $:
$$
2(1.5) + 3(3) = 3 + 9 = 12 ✔ \text{ lies on graph}
$$

✔ Both points lie on the graph.

---

#### 11. Give the geometrical interpretation of $ 5x + 3 = 3x - 7 $ as an equation

i) In one variable
ii) In two variables

---

##### i) In one variable:

The equation $ 5x + 3 = 3x - 7 $ is a linear equation in one variable.

Geometrical interpretation: It represents a point on the number line.

Solve:
$$
5x + 3 = 3x - 7 \Rightarrow 2x = -10 \Rightarrow x = -5
$$

So it represents the point $ x = -5 $ on the real number line.

✔ Geometrical interpretation: A single point on the number line.

---

##### ii) In two variables:

Rewrite the equation in two variables.

Let $ y = 5x + 3 $ and $ y = 3x - 7 $

So the equation $ 5x + 3 = 3x - 7 $ means the two lines $ y = 5x + 3 $ and $ y = 3x - 7 $ intersect.

Geometrical interpretation: The solution is the point of intersection of the two lines.

Solving: $ x = -5 $, then $ y = 5(-5) + 3 = -25 + 3 = -22 $

So intersection point is $ (-5, -22) $

✔ Geometrical interpretation: The point of intersection of two lines.

---

✔ Final Answers Summary:

#### Section A
1. $ m = -\frac{9}{13} $ (not $ -\frac{24}{47} $ — likely error)
2. $ a = 1 $
3. $ a = 1, b = -\sqrt{3}, c = -4 $

#### Section B
4. $ y = -2.5 $
5. $ y = 8x + 7 $
6. $ a = -\frac{1}{4}, b = -\frac{1}{6} $ (not $ \frac{1}{4}, \frac{1}{6} $ — likely sign error)
7. X-axis: $ \left( \frac{4}{7}, 0 \right) $, Y-axis: $ \left( 0, -\frac{4}{3} \right) $

#### Section C
8. $ x = -\frac{19}{11} $ (not 4 — likely error in answer key)
9. Both lines pass through origin and are perpendicular.
10. Both points lie on the graph.
11.
i) A point on the number line ($ x = -5 $)
ii) Point of intersection of two lines

---

Please note: There are discrepancies between the provided answers and correct mathematical solutions in Q1, Q6, and Q8. These appear to be errors in the answer key. The solutions above are mathematically correct.