3. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

 

exercise 7.4 optional exercise co ordinate geometry chapter 7 cbse ncert 10th mathematics

 

 

3. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

 

let the given points be A(6, – 6), B(3, – 7) and C(3, 3). 

let P(h,k) be the centre

using distance formula and the identities

PA = sqrt [ (h-6)^2 + (k+6)^2 ] = sqrt[ (h^2)+(k^2) -12h+12k+72]

PB=sqrt[ (h-3)^2 + (k+7)^2 ] = sqrt[ (h^2)+(k^2) -6h+14k+58]

PC=sqrt[ (h-3)^2 + (k-3)^2 ] = sqrt[ (h^2)+(k^2) -6h-6k+18]

 

Since P is the centre PA=PB=PC

 

PA=PB means on squaring

 [ (h^2)+(k^2) -12h+12k+72] =[ (h^2)+(k^2) -6h+14k+58]

-6h -2k =(-14)--------------------(1)

 

similarly PA = PC gives

 [ (h^2)+(k^2) -12h+12k+72] =[ (h^2)+(k^2) -6h-6k+18]

-6h +18k =(-54) ---------------------------(2)

solving  (1) and (2)

-6h -2k =(-14)--------------------(1)

-6h +18k =(-54) ---------------------------(2)

--------------------------------------------------------------------------- subtracting

        -20 k =(+40)

k= (40)/ (-20)

k = (-2)

substitute in -6h -2k =(-14)

 

-6h  + 4 = (-14)

 

-6h = -14 -4

-6h = (-18)

 

h = (-18) /(-6)

 

h = 3

 

centre is P(h,k) =( 3 , -2 )

 

4. The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

let the given vertices be A(-1,2) and C(3,2)

let B(h,k) be one of the other vertices

using distance formula

AB =sqrt [ (h+1)^2 +(k-2)^2 ] =sqrt[ (h^2) + (k^2) +2h-4k+5 ]

BC = sqrt[(h-3)^2 +(k-2)^2 ] =sqrt[ (h^2) + (k^2) -6h-4k+13 ]

 

AC = sqrt[(3+1)^2 + (2-2)^2] = 4

 

since all sides of a square are equal

 

AB= BC gives

 [ (h^2) + (k^2) +2h-4k+5 ] = [ (h^2) + (k^2) -6h-4k+13 ]

8h = 8

h=1

 

now use pythagoras theorem in triangle ABC to solve for k

 

(AB)^2 + (BC)^2 = (AC)^2

 

  [ (h^2) + (k^2) +2h-4k+5 ] + [ (h^2) + (k^2) -6h-4k+13 ] = 16

2 (h^2) +2(k^2) -4h -8k +2=0

apply h=1

2 +2(k^2) -4 -8k+2 = 0

2(k^2) -8k=0

2k(k-4) =0

k=0 or k=4

so other vertices are (1,0) and (1,4)

 

 

 

 

=================================================

ncert cbse 10th mathematics

 

co ordinate geometry chapter 7

exercise 7.4 optional exercise  

 

Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

 solution 

 

2. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.

solution 

 

3. Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).

solution

4. The two opposite vertices of a square are (–1, 2) and (3, 2). Find the coordinates of the other two vertices.

solution   

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