Question

please solve this problem......

Answer 1

Heya !!

see diagram
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A(3,4) and B(7,7) and C(x,y) and C'(x,y) are collinear points.

AB = √[(7-3)2+(7-4)2] = 5

AC= 10, given

slope of AC = slope of AB = (7-5)/(7-4)2] = 5
AC = 10, given

(y-4)/(x-3) = 3/4 ......(1)
4y - 3x = 7 ........(2)

${ac}^{2} = {10}^{2} = {(y - 4)}^{2} + {(x - 3)}^{2} \\ = ( \frac{4}{3} \times (x - 3)2 + (x - 3)2 \\ using \: (1) \\ \\ = {(x - 3)}^{2} \times \frac{25}{16}$
= x-3 = ±8
= x= +11 or -5
= y = (7+3x)/4 by(2)
= 10 or -2

c = (11,10) and c' = (-5 , -2)
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GLAD HELP YOU.

It helps you,
thank you☻

@vaibhav246

Answer 2

Hey! ! !

Matr :-

Answer :

Given :

☆ A , B and C are three collinear points ,where A = ( 3,4) and B(7,7).

We know distance formula

d = 

So,

Distance between A and B , we get

= 
And
Distance between A and C is 10 unit , So
Distance between B and C  =  10 - 5  =  5 unit 

Let Coordinates of  " C " = (  x  , y  )

So, we get


And


Now we subtract equation 2 from equation 1 , we get

8x  + 6y  = 148

4x  + 3y  = 74

4x  = 74 - 3y 

x  =   , Substitute that value in equation 1 , we get


So,
x  = 
So,

Coordinate of C = (  11 , 10 )                                     ( Ans )