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NCERT Solutions for Class 10 Maths – Coordinate Geometry

1. If the point A (x , y) is equidistant from B (4 , 2) and C (-2 , 4). Find the relation between x and y.

|AB| = (x5)2+(y1)2−−−−−−−−−−−−−−−√

= x2+168x+y2+44y−−−−−−−−−−−−−−−−−−−−−√

|AC| = (x+2)2+(y5)2−−−−−−−−−−−−−−−√

= x2+4+4x+y2+168y−−−−−−−−−−−−−−−−−−−−−√

Since, |AB| = |AC|

x2+y28x4y+20 = x2+y2+4x8y+20

      8y – 4y = 4x + 8x

      y = 3x

 2. Find the perimeter of a triangle with vertices (0, 8), (0, 0) and (6, 0)

Let the vertices of the be P(0, 8), Q(0, 0) and R(6, 0)

∴ PQ = (0)2+(8)2−−−−−−−−−−√=64−−√=8

∴ QR = (6)2+(0)2−−−−−−−−−√=36−−√=6

∴ RP = (6)2+(8)2−−−−−−−−−−√=100−−−√=10

∴  Perimeter of = 8 + 6 + 10 = 24 units

3.Find the coordinates of the perpendicular bisector of the line segment joining the points P (1, 4) and Q (2, 3), cuts the y-axis.

Here, O (0, y), P (1, 4) and Q (2, 3)

AO = BO

(01)2+(y4)2−−−−−−−−−−−−−−−√ = (02)2+(y3)2−−−−−−−−−−−−−−−√

1+y2+168y=4+y2+96y

   8y – 6y = 17 -13

   2y = 4

y = 2

∴The required point is (0, 2)

4. The points which divides the line segment joining the points (5, 8) and (1, 2) in ratio 1:2 internally lies in which quadrant?

Now C(1+103,2+163)

∴ C(113,6)

Capture

Since, C(113,6) lies in IV quadrant.

5. If A (x2,3) is the mid-point of the line segment joining the points B(4,1) and C(8,7). Find the value of x.

4+82=a2

24 = 2a

a = 12

 

6. A line intersects the y-axis at point A and B respectively. If (3, 4) is the mid-point of AB. Find the coordinates of A and B.

Here, x+02=3 x=6

0+y2=4 x=8

2

∴ The coordinates of A and B are (0, 8) and (6, 0).

7. Find the fourth vertex S of a parallelogram PQRS whose three vertices are P (-3, 8), Q (5, 4) and R (6, 2).

Here, 3+62=5+x2x=1

8+22=4+x2x=3

∴The fourth vertices S of a parallelogram PQRS is S(-1, 3)

 8. Find the coordinates of the point which is equidistant from the three vertices of the

POQ.

3

Point equidistant from the three vertices of  a right angle triangle is the mid-point of hypotenuse.

(4x+02,0+4y2)(2x,2y)

 

9. Find the area of a triangle with vertices (x, y + z), (y, z + x) and (z, x + y).

Area of the required

=12|x(z+xxy)+y(x+yyz)+(y+zzx)|

= 12|x(zy)+y(xz)+z(ya)|

= 12|0| = 0.

10. Find the area of a triangle with vertices P (2, 0), Q (6, 0) and R (5, 4)

Area of the required PQR

= 12|2(02)+6(40)+5(00)|

= 12|244|=10 Sq. units.

11. If the points O (0, 0), P (3, 4), Q (x, y) are collinear, then write the relation between x and y.

Since, O (0, 0), P (3, 4), Q (x, y) are collinear.

Area of a triangle formed by these points vanishes

∴ 12|0(4y)+3(y0)+x(04)|=0

y2x=02x=y

 

12. Find the coordinates of the point O dividing the line segment joining the point P (6, 5)

and Q(2, 10) in the ratio 2:1.

4

x=2(2)+1(6)5=4+65=2

and y=2(10)+1(5)5=20+55=5

∴ O(2, 5)

13. The line joining A(4, 3) and B(-3, 6) meets y-axis at C. At what ratio dose C divides the line segment AB?

Since,R lies on y-axis, let C be (0, y)

5

Suppose that the required ratio be z:1

x=3z+4z+10=3z+4z=43

∴ The required ratio is 4 : 3.

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2017-09-28T13:19:33+00:00 Categories: CBSE|0 Comments
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