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**Q1. Points P (5, âˆ’ 3) is one of the two points of trisection of the line segment joining the points A (7, âˆ’ 2) and B (1, âˆ’ 5) near to A. Find the coordinates of the other point of trisection.**

**Sol.**

âˆ´ other point Q is the mid point of PB

Thus, the point Q is (3, âˆ’ 4)

**Q2. Find the area of the quadrilateral ABCD whose vertices are A (1, 0), B (5, 3), C (2, 7) and D (âˆ’ 2, 4).**

**Sol. **Area of Î”ABC

Area of Î” ACD

**Q3. Points P, Q, R and S, in this order, divide a line segment joining A (2, 6), B (7, âˆ’ 4) in five equal parts. Find the coordinates of P and R.**

**Sol.**

âˆµ P, Q, R and S divide AB in five equal parts.

âˆ´ AP = PQ = QR = RS = SB

Now, P divides AB in the ratio 1 : 4

âˆ´ Coordinates of P are:

Again, R divides AB in the ratio 3 : 2

âˆ´ Coordinates of R are:

**Q4. A (âˆ’ 4, âˆ’ 2), B (âˆ’ 3, âˆ’ 5), C (3, âˆ’ 2) and D (2, k) are the vertices of a quad. ABCD. Find the value of k, if the area of the quad is 28 sq. units.**

**Sol. **Area of quad ABCD = 28 sq. units

âˆ´ [ar (Î” ABD)] + [ar (Î” BCD)] = 28 sq. units

**Q5. Find the point on y-axis which is equidistant from the points (5, âˆ’ 2) and (âˆ’ 3, 2).**

**Sol. **Let the required point be P (0, y) Ã¤ The given points are A (5, âˆ’ 2) and B (âˆ’ 3, 2)

âˆ´ PA = PB

â‡’ PA^{2} = PB^{2}

âˆ´(5 âˆ’ 0)^{2} + (âˆ’ 2 âˆ’ y^{)2} =(âˆ’ 3 âˆ’ 0)^{2} + (2 âˆ’ y)^{2}

â‡’ 5^{2} + (âˆ’ 2 âˆ’ y)^{2} = (âˆ’ 3)^{2} + (2 âˆ’ y)^{2}

â‡’ 25 + 4 + y^{2} + 4y = 9 + 4 + y^{2} âˆ’ 4y

â‡’ 25 + 4y = 9 âˆ’ 4y

â‡’ 8 y = âˆ’ 16 â‡’ y = âˆ’ 2

Thus, the required point is (0, âˆ’ 2)

**Q6. Find the point on y-axis which is equidistant from (âˆ’ 5, 2) and (9, âˆ’ 2).**

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

The given points are A (âˆ’ 5, 2) and B (9, âˆ’ 2)

âˆ´ AP = BP

â‡’ âˆ’ 4y âˆ’ 4y = 81 + 4 âˆ’ 4 âˆ’ 25

â‡’ âˆ’ 8y =56

â‡’

âˆ´ The required point = (0, âˆ’7)

**Q7. Find the value of x for which the distance between the points P (4, âˆ’ 5) and Q (12, x) is 10 units. **

**Sol. **The given points are P (4 âˆ’ 5) and Q (12, x) such that PQ = 10

â‡’ (12 âˆ’ 4)^{2} + (x + 5)^{2} =10^{2}

â‡’ 82 + (x + 5)^{2} = 100

â‡’ 64 + x^{2} + 25 + 10x = 100

â‡’ x^{2} + 10x âˆ’ 11 = 0

â‡’ (x âˆ’ 1) (x + 11) = 0

â‡’ x = 1 or x = âˆ’11

**Q8. Find the relation between x and y if the points (2, 1), (x, y) and (7, 5) are collinear.**

**Sol. **Here, x_{1} = 2, y_{1} = 1

x_{2} = x, y_{2} = y

x_{3} = 7, y_{3} = 5

âˆ´ Area of triangle

which is the required relation.

**Q9. If A (âˆ’ 2, 4), B (0, 0) and C (4, 2) are the vertices of Î” ABC, then find the length of the median through the vertex A.**

**Sol. **âˆµ AD is the median on BC

âˆ´ D is the mid-point of BC.

â‡’ Coordinates of D are:

Now, the length of the median

**Q10. If the points A (4, 3) and B (x, 5) are on the circle with the centre O (2, 3), find the value of x.**

**Sol.** Let O (2, 3) be the centre of the circle.

âˆ´ OA = OB â‡’ OA^{2} = OB^{2}

â‡’ (4 âˆ’ 2)^{2} + (3 âˆ’ 3)^{2} =(x âˆ’ 2)^{2} + (5 âˆ’ 3)^{2}

â‡’ 2^{2} =(x âˆ’ 2)^{2} + 2^{2}

â‡’ (x âˆ’ 2)^{2} = 0

â‡’ x âˆ’2 = 0

â‡’ x = 2

Thus, the required value of x is 2.

**Q11. If the vertices of a Î” are (2, 4), (5, k) and (3, 10) and its area is 15 sq. units, then find the value of â€˜kâ€™.**

**Sol. **The area of the given Î”

But ar (Î”) = 1 5 [given]

â‡’ âˆ’ k = 30 âˆ’ 22 = 8

â‡’ k = âˆ’ 8

**Q12. The vertices of a triangle are: (1, k), (4, âˆ’ 3), (âˆ’ 9, 7) and its area is 15 sq. units. Find the value of k.**

**Sol. **Area of the given triangle

**Q13. Find the area of a Î”ABC whose vertices are A (âˆ’ 5, 7), B (âˆ’ 4, âˆ’ 5) and C (4, 5).**

**Sol. **Here, x_{1} = âˆ’ 5, y_{1} = 7

x_{2} = âˆ’ 4, y_{2} = âˆ’ 5

x_{3} =4, y_{3} = 5

**âˆ´ The required ar (Î” ABC) = 53 sq. units.**

**Q14. Find the value of k such that the points (1, 1), (3, k) and (âˆ’ 1, 4) are collinear.**

**Sol.** For the three points, to be collinear, the area of triangle formed by them must be zero.

âˆ´ Area of triangle = 0

â‡’ k = - 2

**Q15. For what value of p, the points (âˆ’ 5, 1), (1, p) and (4, âˆ’ 2) are collinear?**

**Sol. **Since the points are collinear,

âˆ´ The area of the Î” formed by these points must be zero.

**Q16. For what value of p, are the points (2, 1), (p, âˆ’ 1) and (âˆ’ 1, 3) collinear? **

**Sol. **âˆµ The given points are collinear.

âˆ´ The area of a triangle formed by these points must be zero. i.e.,Area of triangle = 0

**Q17. Find the ratio in which the line 3x + 4y âˆ’ 9 = 0 divides the line segment joining the points (1, 3) and (2, 7).**

**Sol.** Let the ratio be k : 1.

â‡’ 6k + 3 + 28k + 12 âˆ’ 9k + 9 = 0

â‡’ (6k + 28k âˆ’ 9k) + (3 + 12 âˆ’ 9) = 0

â‡’ 25k + 6 = 0

â‡’

âˆ´ The required ratio is âˆ’ 6 : 25 or 6 : 25

**Q18. If the point P (x, y) is equidistant from the points A (3, 6) and B (âˆ’ 3, 4), prove that 3x + y âˆ’ 5 = 0.**

**Sol.**

âˆµ P is equidistant from A and B.

Q19. The coordinates of A and B are (1, 2) and (2, 3). If P lies on AB, then find the coordinates of P such that:

**Sol.**

**Q20. If A (4, âˆ’ 8), B (3, 6) and C (5, âˆ’ 4) are the vertices of a Î” ABC, D(4, 1) is the mid-point of BC and P is a point on AD joined such that AP/PD = 2 , find the coordinates of P.**

**Sol. âˆµ D is the mid-point of B**

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