Worksheet for Chapter 7 on Co-ordinate Geometry covering various problems on points, lines, and triangles.

Let's solve each of these coordinate geometry problems step by step with explanations.

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1. In which quadrant does the point (–10, 2) lie?

- The x-coordinate is negative (–10), and the y-coordinate is positive (2).
- This lies in Quadrant II.
✔ Answer: Quadrant II

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2. (0, 2) and (0, –5) are the coordinates of two points lying on ________ axis.

- Both points have x = 0 → they lie on the y-axis.
✔ Answer: y-axis

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3. Find the distance of a point P(x, y) from the origin (0, 0).

- Use the distance formula:
$$
\text{Distance} = \sqrt{(x - 0)^2 + (y - 0)^2} = \sqrt{x^2 + y^2}
$$
✔ Answer: $\sqrt{x^2 + y^2}$

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4. What are the coordinates of the midpoint of the line joining the points (6, –2) and (4, 8)?

- Midpoint formula:
$$
\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
$$
$$
= \left( \frac{6 + 4}{2}, \frac{-2 + 8}{2} \right) = \left( \frac{10}{2}, \frac{6}{2} \right) = (5, 3)
$$
✔ Answer: (5, 3)

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5. What are the coordinates of the centroid of triangle formed by the points (–7, 6), (8, 5), (2, –2)?

- Centroid formula:
$$
\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)
$$
$$
= \left( \frac{-7 + 8 + 2}{3}, \frac{6 + 5 + (-2)}{3} \right) = \left( \frac{3}{3}, \frac{9}{3} \right) = (1, 3)
$$
✔ Answer: (1, 3)

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6. What is the x-coordinate of the point which divides the line joining (1, 2) and (2, 3) in the ratio 4 : 3?

- Section formula (internal division):
$$
x = \frac{m x_2 + n x_1}{m + n}, \quad \text{where } m:n = 4:3
$$
$$
x = \frac{4 \cdot 2 + 3 \cdot 1}{4 + 3} = \frac{8 + 3}{7} = \frac{11}{7}
$$
✔ Answer: $\frac{11}{7}$

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7. Find the coordinates of the point which divides the line joining (–4, 0) and (0, 6) in the ratio 1 : 3.

- Using section formula:
$$
x = \frac{1 \cdot 0 + 3 \cdot (-4)}{1 + 3} = \frac{0 - 12}{4} = -3
$$
$$
y = \frac{1 \cdot 6 + 3 \cdot 0}{4} = \frac{6}{4} = 1.5 = \frac{3}{2}
$$
✔ Answer: $(-3, \frac{3}{2})$

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8. Find the third vertex of a triangle if two of its vertices are (–1, 4) and (5, 2) and centroid is (0, –3).

- Let the third vertex be $(x, y)$.
- Centroid:
$$
\left( \frac{-1 + 5 + x}{3}, \frac{4 + 2 + y}{3} \right) = (0, -3)
$$
Solve:
$$
\frac{4 + x}{3} = 0 \Rightarrow 4 + x = 0 \Rightarrow x = -4
$$
$$
\frac{6 + y}{3} = -3 \Rightarrow 6 + y = -9 \Rightarrow y = -15
$$
✔ Answer: $(-4, -15)$

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9. The length of the line AB, where A(1, 0) and B(5, 3) is ________ units.

- Distance formula:
$$
AB = \sqrt{(5 - 1)^2 + (3 - 0)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5
$$
✔ Answer: 5 units

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10. Find the length of the line AB, where coordinates of points A and B are (2, 7) and (–2, 4).

- Distance:
$$
AB = \sqrt{(-2 - 2)^2 + (4 - 7)^2} = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5
$$
✔ Answer: 5 units

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11. What is the distance of the point (8, –2) from the origin?

- Distance from origin:
$$
\sqrt{8^2 + (-2)^2} = \sqrt{64 + 4} = \sqrt{68} = 2\sqrt{17}
$$
✔ Answer: $2\sqrt{17}$ units

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12. Find the centroid of a triangle whose vertices are (–2, –3), (–1, 0), and (7, –6)

- Centroid:
$$
x = \frac{-2 + (-1) + 7}{3} = \frac{4}{3}, \quad y = \frac{-3 + 0 + (-6)}{3} = \frac{-9}{3} = -3
$$
✔ Answer: $\left(\frac{4}{3}, -3\right)$

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13. What is the distance between points (–3, 2) and (1, –2)?

- Distance:
$$
\sqrt{(1 - (-3))^2 + (-2 - 2)^2} = \sqrt{(4)^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}
$$
✔ Answer: $4\sqrt{2}$ units

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14. What is the area of triangle ABC, whose vertices are A(x₁, y₁), B(x₂, y₂), C(x₃, y₃)?

- Area formula:
$$
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
$$
✔ Answer: $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

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15. Find the coordinates of the point dividing the points (3, 5) and (7, 9) in the ratio 2 : 3.

- Section formula:
$$
x = \frac{2 \cdot 7 + 3 \cdot 3}{2 + 3} = \frac{14 + 9}{5} = \frac{23}{5}
$$
$$
y = \frac{2 \cdot 9 + 3 \cdot 5}{5} = \frac{18 + 15}{5} = \frac{33}{5}
$$
✔ Answer: $\left(\frac{23}{5}, \frac{33}{5}\right)$

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16. Find the distance of the point (0, 2) from the midpoint of the line joining (4, 10) and (2, 2).

- First, find midpoint:
$$
\left( \frac{4 + 2}{2}, \frac{10 + 2}{2} \right) = (3, 6)
$$
- Distance from (0, 2) to (3, 6):
$$
\sqrt{(3 - 0)^2 + (6 - 2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5
$$
✔ Answer: 5 units

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17. Point Q lies on the line joining P(–3, 2) and origin in such a way that OP = OQ. What will be the coordinates of Q?

- OP is the distance from origin to P(–3, 2):
$$
OP = \sqrt{(-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13}
$$
- Since Q lies on the line joining P and origin, and OP = OQ, but Q must be on the same line, then either:
- Q is at P → but that would mean OQ = OP, so Q = P?
- But "OP = OQ" suggests that Q is at same distance from origin as P, but on the line joining P and origin.

But since it says "lies on the line joining P and origin", and OP = OQ, then:

- If Q is on the ray from origin through P, then Q could be in the opposite direction?

Wait — let's think carefully.

The line joining P(–3, 2) and origin (0, 0). Any point on this line can be written as $k(-3, 2)$ for scalar $k$.

Let Q = $k(-3, 2) = (-3k, 2k)$

Then OQ = distance from origin = $\sqrt{(-3k)^2 + (2k)^2} = \sqrt{9k^2 + 4k^2} = \sqrt{13k^2} = |k|\sqrt{13}$

We want OQ = OP = $\sqrt{13}$

So:
$$
|k|\sqrt{13} = \sqrt{13} \Rightarrow |k| = 1 \Rightarrow k = \pm 1
$$

- If $k = 1$: Q = (–3, 2) = P → valid
- If $k = -1$: Q = (3, –2)

Now, does Q lie on the line joining P and origin? Yes — because it's the extension beyond the origin.

But the problem says: "Point Q lies on the line joining P and origin" — this includes both directions.

Also, OP = OQ → both are $\sqrt{13}$

So possible points: (–3, 2) or (3, –2)

But if Q = P, then it's trivial. Probably they want the other point symmetric about origin.

Thus, Q is the reflection of P across the origin → (3, –2)

✔ Answer: (3, –2)

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18. State which of the two lines is longer:

(i) Line joining (0, 1) and (4, –2)
(ii) Line joining (1, 2) and (6, 4)

#### (i) Length:
$$
\sqrt{(4 - 0)^2 + (-2 - 1)^2} = \sqrt{16 + 9} = \sqrt{25} = 5
$$

#### (ii) Length:
$$
\sqrt{(6 - 1)^2 + (4 - 2)^2} = \sqrt{25 + 4} = \sqrt{29} \approx 5.385
$$

Since $\sqrt{29} > 5$, line (ii) is longer.

✔ Answer: Line (ii) is longer

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✔ Final Answers Summary:

1. Quadrant II
2. y-axis
3. $\sqrt{x^2 + y^2}$
4. (5, 3)
5. (1, 3)
6. $\frac{11}{7}$
7. $(-3, \frac{3}{2})$
8. $(-4, -15)$
9. 5 units
10. 5 units
11. $2\sqrt{17}$ units
12. $\left(\frac{4}{3}, -3\right)$
13. $4\sqrt{2}$ units
14. $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$
15. $\left(\frac{23}{5}, \frac{33}{5}\right)$
16. 5 units
17. (3, –2)
18. Line (ii) is longer

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