Class IX Mathematics worksheet on linear equations in two variables, featuring problems from Section A to C with 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:

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

Yes, and we substituted correctly.

But perhaps there’s a misprint. Let’s check again.

Wait — maybe it was supposed to be $ y - \frac{3}{2} = m(x - 7) $? But no, the equation is as written.

Alternatively, perhaps the point is not $ (7, -3) $? Or maybe the equation is different?

Wait — actually, let's see: If the answer is $ m = -\frac{24}{47} $, then let's verify whether this value satisfies the condition.

Try plugging back:

If $ m = -\frac{24}{47} $, then:

$$
y - \frac{3}{2} = -\frac{24}{47}(x - \frac{1}{2})
$$

At $ x = 7 $:

$$
y = \frac{3}{2} - \frac{24}{47}(7 - 0.5) = \frac{3}{2} - \frac{24}{47}(6.5)
= \frac{3}{2} - \frac{24}{47} \cdot \frac{13}{2}
= \frac{3}{2} - \frac{312}{94}
= \frac{3}{2} - \frac{156}{47}
$$

Convert $ \frac{3}{2} = \frac{141}{94} $, $ \frac{156}{47} = \frac{312}{94} $

So $ y = \frac{141 - 312}{94} = \frac{-171}{94} \approx -1.82 $, not $ -3 $

So clearly, $ m = -\frac{24}{47} $ is not correct for this point.

But our calculation gives $ m = -\frac{9}{13} $. So either the given answer is wrong, or the equation is different.

Wait — perhaps the equation is $ y - \frac{3}{2} = m(x - \frac{7}{2}) $? That would make more sense.

But as per the image, it says $ y - \frac{3}{2} = m(x - \frac{1}{2}) $

So unless there’s a typo, the correct value is:

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

But since the answer key says $ m = -\frac{24}{47} $, perhaps the point is different? Or maybe the equation is $ y - 3 = m(x - 7) $? That’s a possibility.

Wait — perhaps the equation is $ y - \frac{3}{2} = m(x - 7) $? Then:

$ -3 - 1.5 = m(7 - 7) \Rightarrow -4.5 = m(0) $ → impossible.

No.

Alternatively, maybe the point is $ (7, -3) $, but the equation is $ y - \frac{3}{2} = m(x - \frac{7}{2}) $? Try that.

Then:

$ -3 - 1.5 = m(7 - 3.5) \Rightarrow -4.5 = m(3.5) \Rightarrow m = -\frac{4.5}{3.5} = -\frac{9}{7} $

Still not matching.

Alternatively, maybe the equation is $ y + 3 = m(x - 7) $? But that’s not what’s written.

Given the discrepancy, and assuming the problem is correct as stated, then our calculation stands:

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

But since the expected answer is $ -\frac{24}{47} $, let's suppose the equation is:

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

and the point is $ (7, -3) $, then:

We already did:

Left side: $ -3 - 1.5 = -4.5 $

Right side: $ m(7 - 0.5) = m(6.5) $

So $ m = -4.5 / 6.5 = -9/13 $

So unless the point is different, this can't be.

Wait — could it be $ y - \frac{3}{2} = m(x - \frac{7}{2}) $? Try:

$ x = 7 $, so $ x - 3.5 = 3.5 $

$ y = -3 $, so $ y - 1.5 = -4.5 $

So $ m = -4.5 / 3.5 = -9/7 $

Still not matching.

Perhaps the equation is $ y - 3 = m(x - 7) $? Then at $ x = 7 $, $ y = 3 $, but our point is $ (7, -3) $, so doesn’t work.

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

$ -3 - 1.5 = -4.5 $, $ x - 3.5 = 3.5 $, so $ m = -4.5 / 3.5 = -9/7 $

No.

I think the only possibility is that the given answer $ m = -\frac{24}{47} $ is incorrect, or the point or equation is different.

But since the user has provided the answer as $ m = -\frac{24}{47} $, and we're to explain, perhaps there’s a typo in the question.

Alternatively, maybe the equation is:

$$
y - \frac{3}{2} = m(x - \frac{7}{2})
$$

and the point is $ (7, -3) $, then:

As above, $ m = -9/7 $

Not matching.

Wait — maybe the equation is $ y - \frac{3}{2} = m(x - \frac{1}{2}) $, and the point is $ (7, -3) $, but the answer is indeed $ -9/13 $, so likely the provided answer is wrong.

So I will go with:

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

But since the key says $ -\frac{24}{47} $, perhaps the point is $ (7, -3) $, but the equation is different.

Wait — another idea: Maybe the equation is $ y - \frac{3}{2} = m(x - \frac{1}{2}) $, but the point is $ (7, -3) $, and we need to find $ m $, which we did.

So unless there's a mistake in the problem statement, the answer should be $ -\frac{9}{13} $.

But let’s move on, and assume the answer key might have an error.

---

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

Substitute $ x = 2 $, $ y = 3 $ into the equation:

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

✔ Answer: $ \boxed{a = 1} $

---

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

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

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 slope is 0.

It is 2.5 units below the x-axis → $ y = -2.5 $

So the equation is:

$$
\boxed{y = -2.5} \quad \text{or} \quad \boxed{y + 2.5 = 0}
$$

---

#### 5. For the first km, the fare is Rs 15 and for the 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 $ = distance in km

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

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

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

Answer given: $ y = 15 + 8(x - 1) $ → which is correct, but simplified form is better.

✔ So final equation: $ \boxed{y = 8x + 7} $

---

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

We substitute both points into the equation:

Point (2, 3):
$$
a(2) + b(3) + 1 = 0 \Rightarrow 2a + 3b = -1 \quad \text{(1)}
$$

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

Now plug $ a = -\frac{1}{4} $ into (1):

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

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

Wait — let's check:

With $ a = -\frac{1}{4}, b = -\frac{1}{6} $, test (2,3):

$ 2(-1/4) + 3(-1/6) + 1 = -0.5 - 0.5 + 1 = 0 $ → OK

Test (4,0): $ 4(-1/4) + 0 + 1 = -1 + 1 = 0 $ → OK

So correct.

But answer key says $ a = \frac{1}{4}, b = \frac{1}{6} $

Try that: $ \frac{1}{4}x + \frac{1}{6}y + 1 = 0 $

At (2,3): $ 0.5 + 0.5 + 1 = 2 \ne 0 $

So incorrect.

So the correct values are:

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

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

Wait — perhaps the equation is $ ax + by = -1 $? No, it's $ ax + by + 1 = 0 $

So the correct answer is:

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

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

Alternatively, maybe the equation is $ ax + by = 1 $? But no, it's $ ax + by + 1 = 0 $

So I’ll stick with:

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

---

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

X-axis intercept: Set $ y = 0 $

$$
7x = 4 \Rightarrow x = \frac{4}{7}
\Rightarrow \text{Point: } \left( \frac{4}{7}, 0 \right)
$$

Y-axis intercept: Set $ x = 0 $

$$
-3y = 4 \Rightarrow y = -\frac{4}{3}
\Rightarrow \text{Point: } \left( 0, -\frac{4}{3} \right)
$$

But the answer key says:

- X axis: $ \left( \frac{4}{7}, 0 \right) $ ✔
- Y axis: $ \left( 0, -\frac{4}{3} \right) $ → but key says $ \left( 0, -\frac{1}{3} \right) $? Wait, no — it says $ (0, -\frac{1}{3}) $, but that's wrong.

Wait — the image says: "Y axis (0, -1/3)" — but that’s incorrect.

Correct is $ \left( 0, -\frac{4}{3} \right) $

So answer key has a typo.

✔ Correct answers:

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

---

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:

$$
105 \left( \frac{3x+2}{7} + \frac{4x+1}{5} \right) = 105 \cdot \frac{2(2x+1)}{3}
$$

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

Compute:

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

Right:
$$
70(2x+1) = 140x + 70
$$

Now:

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

But the answer key says $ x = 4 $

Check: Plug $ x = 4 $ into original:

Left:
$$
\frac{3(4)+2}{7} = \frac{14}{7} = 2 \\
\frac{4(4)+1}{5} = \frac{17}{5} = 3.4 \\
\text{Sum: } 2 + 3.4 = 5.4
$$

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

5.4 ≠ 6 → not equal.

So $ x = 4 $ is not correct.

Our solution: $ x = -\frac{19}{11} \approx -1.727 $

But let’s double-check calculations.

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

Right:
$$
35 × 2(2x+1) = 70(2x+1) = 140x + 70
$$

So:

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

✔ Correct.

But answer key says $ x = 4 $ — must be wrong.

So correct answer: $ \boxed{x = -\frac{19}{11}} $

---

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

- $ y = x $: passes through origin, slope 1 → diagonal upward
- $ y = -x $: passes through origin, slope -1 → diagonal downward

They intersect at origin $ (0,0) $

They are symmetric about x-axis and y-axis.

They form an 'X' shape, perpendicular to each other.

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

✔ So: The graphs intersect at the origin and are perpendicular.

---

#### 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 $ → $ (6, 0) $
- Y-intercept: $ x = 0 $ → $ 3y = 12 $ → $ y = 4 $ → $ (0, 4) $

Plot these and draw the line.

Now check points:

Point (4.5, 1):

$ 2(4.5) + 3(1) = 9 + 3 = 12 $ → YES, lies on graph.

Point (1.5, 3):

$ 2(1.5) + 3(3) = 3 + 9 = 12 $ → YES, 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:

This is a linear equation in one variable $ x $.

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

Geometrical interpretation: On a number line, it represents a single point $ x = -5 $.

---

(ii) In two variables:

Rewrite the equation as:

$$
5x + 3 = 3x - 7 \Rightarrow 5x - 3x + 3 + 7 = 0 \Rightarrow 2x + 10 = 0
$$

But this is still in one variable.

To interpret in two variables, we treat it as:

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

Then the equation $ 5x + 3 = 3x - 7 $ represents the point of intersection of the two lines $ y = 5x + 3 $ and $ y = 3x - 7 $

So geometrically, it represents the intersection point of two straight lines.

The solution $ x = -5 $ corresponds to $ y = 5(-5) + 3 = -25 + 3 = -22 $

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

✔ Geometrical interpretation: The solution is the point where the two lines $ y = 5x + 3 $ and $ y = 3x - 7 $ intersect.

---

✔ Final Summary of Answers:

| Q | Answer |
|---|--------|
| 1 | $ m = -\frac{9}{13} $ (but key says $ -\frac{24}{47} $ — likely error) |
| 2 | $ a = 1 $ |
| 3 | $ a = 1, b = -\sqrt{3}, c = -4 $ |
| 4 | $ y = -2.5 $ |
| 5 | $ y = 8x + 7 $ |
| 6 | $ a = -\frac{1}{4}, b = -\frac{1}{6} $ (key has sign error) |
| 7 | X-axis: $ (\frac{4}{7}, 0) $, Y-axis: $ (0, -\frac{4}{3}) $ (key has typo) |
| 8 | $ x = -\frac{19}{11} $ (key says $ x = 4 $ — incorrect) |
| 9 | Lines intersect at origin and are perpendicular |
| 10 | Both points lie on the graph |
| 11 | i) Point $ x = -5 $ on number line<br>ii) Intersection point of two lines |

---

⚠️ Note: Several answers in the provided key appear to be incorrect. Please verify the original problem statements.