CHAPTER 7 Coordinate geometry solution

 

COORDINATE GEOMETRY SOLUTION AND EXPLANATIONS FOR CLASS X

Revision 

All the distances are measured from the origin "O" (0,0) 

In the first quadrant, x and y distances are positive, in the second quadrant, x is negative and y is positive, in the third quadrant, both x and y are negative, and in the fourth quadrant, x is positive and y is negative. 

Distance Formula

Distance of A from the origin is x1

Distance of B from origin = x2

AB = x2-x1

AQ = y1

BP = y2

The distance between Q and P is obtained by using the Pythagorean formula

QP is the required distance mentioned in the figure.

This is not a quadrilateral.

The given coordinates form a parallelogram

Solve the (iv) part similarly. 

Solution 7:

Solution 8: Given points are P and Q

Solution 9 

Solution 10:

Q7 solution

If A and B are the coordinates of the diameter 

Then the center will be the midpoint of AB.

Let the coordinates of A be (x,y)

2 = (x+1)/2 

4 = x+1

x =3

3 = (y+4)/2

6 = y+4

y = 20

Therefore, the coordinate of A(x, y) = (3, 2).

Solution 8:

If AP = 3/7 AB

BP = 1 - 3/7AB

     = 4/7 AB

AP: PB = 3:4

The coordinates of A and B are (-2, -2) and (2, -4)

Hence, P(x,y) 

x = (3*2 + 4*(-2))/(3+4) = -2/7

y = (3*(-4) + 4*(9-2))/(4+3) = -20/7

P(x,y) = (-2/7, -20/7)

; 

Solution 9

Let the P(a,b), Q(c,d), and R(e,f) be the coordinates of three points  that divide AB into four equal parts

Then, AP:PQ:QR: RB = 1:1:1:1 

To find the coordinate of P(a,b), consider the ratio AP:PB = 1:3

a= (1*2 + 3*(-2))/(4)

a = -4/4 = -1

b = (1*8 + 3*2)/(4) = 14/4 = 7/2

P(a, b) = (-1, 7/2)

For the coordinates of Q (c,d)

AQ: QB = 2:2 = 1:1

c = (-2 + 2)/4 = 0

d = (2+8)/4 = 5/2

Q(c, d) = (0, 5/2)

For the coordinates of R(e, f)

AR: RB = 3:1

e = (3*2 + 1*(-2))/4

    = 4/4 = 1

f = (3*8 + 1*2)/4 

   = 13/2 

R(e, f) = (1, 13/2)

Solution 10:

Let the vertices of a rhombus be A(3, 0), B(4, 5), C(– 1, 4) and D(– 2, – 1)

Area of a Rhombus = Product of diagonals / 2