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# Short Answer Type Questions(Part- 1)- Coordinate Geometry Class 10 Notes | EduRev

## Class 10 : Short Answer Type Questions(Part- 1)- Coordinate Geometry Class 10 Notes | EduRev

The document Short Answer Type Questions(Part- 1)- Coordinate Geometry Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.
All you need of Class 10 at this link: Class 10

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‡’ PA2 = PB2
âˆ´(5 âˆ’ 0)2 + (âˆ’ 2 âˆ’ y)2 =(âˆ’ 3 âˆ’ 0)2 + (2 âˆ’ y)2
â‡’ 52 + (âˆ’ 2 âˆ’ y)2 = (âˆ’ 3)2 + (2 âˆ’ y)2
â‡’ 25 + 4 + y2 + 4y = 9 + 4 + y2 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 =102
â‡’ 82 + (x + 5)2 = 100
â‡’ 64 + x2 + 25 + 10x = 100
â‡’ x2 + 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, x1 = 2, y1 = 1
x2 = x, y2 = y
x3 = 7, y3 = 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‡’ OA2 = OB2
â‡’ (4 âˆ’ 2)2 + (3 âˆ’ 3)2 =(x âˆ’ 2)2 + (5 âˆ’ 3)2
â‡’ 22 =(x âˆ’ 2)2 + 22
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, x1 = âˆ’ 5, y1 = 7
x2 = âˆ’ 4, y2 = âˆ’ 5
x3 =4,  y3 = 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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