The document Very Short Answer Type Questions- Coordinate Geometry Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.

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**Q1. Find a point on the y-axis equidistant from (− 5, 2) and (9, − 2).**

**Sol. **Let the required point on the y-axis be P (0, y)

∴ PA = PB

⇒ y^{2} − y^{2} − 4y − 4y = 81 + 4 − 4 − 25

⇒ − 8y = 85 − 29

⇒ − 8y = 56

⇒

∴ The required point is (0, −7).

**Q2. Find a point on x-axis at a distance of 4 units from the point A (2, 1).**

**Sol. **Let the required point on x-axis be P (x, 0).

∴ PA = 4

⇒ x^{2} − 4x + 4 + 1 = 4^{2} = 16

⇒ x^{2} − 4x + 1 + 4 − 16 = 0

⇒ x^{2} − 4x − 11 = 0

⇒

**Q3. Find the distance of the point (3, − 4) from the origin.**

**Sol. **The coordinates of origin (0, 0).

∴ Distance of (3, − 4) from the origin

**Q4. For what value of x is the distance between the points A (− 3, 2) and B (x, 10) 10 units?**

**Sol. **The distance between A (− 3, 2) and B (x, 10)

⇒ (x + 3)^{2} + (8)^{2} = 10^{2}

⇒ (x + 3)^{2} = 10^{2} − 8^{2}

⇒ (x + 3)^{2} = (10 − 8) (10 + 8) = 36

⇒

For +ve sign, x =6 − 3 = 3

For −ve sign, x = − 6 − 3 = − 9

**Q5. Find a point on the x-axis which is equidistant from the points A (5, 2) and B (1, − 2). Sol. **The given points are: A (5, 2) and B (1, − 2) Let the required point on the x-axis be C (x, 0).

Since, C is equidistant from A and B.

∴ AC = BC

∴ The required point is (0, 3).**Q6. Establish the relation between x and y when P (x, y) is equidistant from the points A (− 1, 2) and B (2, − 1).**

**Sol. **∵ P is equidistant from A and B

∴ PA = PB

which is the required relation.

**Q7. Find a relation between x and y such that the point P (x, y) is equidistant from the points A (−5, 3) and B (7, 2) **

**Sol.** Since, P (x, y) is equidistant from A (−5, 3) and B (7, 2)

∴ AP = BP

**Q8. Show that the points (7, − 2), (2, 3) and (− 1, 6) are collinear.**

**Sol. **Here, the vertices of a triangle are (7, − 2), (2, 3) and (− 1, 6)

∴ Area of the triangle

Since area of triangle = 0

∴ The vertices of the triangle are collinear.

Thus, the given points are collinear.

**Q9. Find the distance between the points **

**Sol.** Distance between is given by

**Q10. If the mid point of the line joining the points P (6, b − 2) and Q (− 2, 4) is (2, − 3), find the value of b.**

**Sol. Here, P (6, b − 2) and Q (− 2, 4) are the given points. ∴ Mid point of PQ is given by:**

**Q11. In the given figure, ABC is a triangle. D and E are the mid points of the sides BC and AC respectively. Find the length of DE. Prove that **

**Sol.** Coordinates of the mid point of BC are:

Coordinates of the mid point of AC are:

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