Question

5. If coordinates of A and B are (3, 4) and (6. 8) respectively and C and D are two points on AB such that
AC = CD = DB, then find the coordinates of points C and D.
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Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

$(4, \frac{16}{3}) and (5,\frac{20}{3} )$ are the coordinates of C and D .

Step-by-step explanation:

Explanation:

Given , coordinate of A (3,4) and coordinate of B(6,8)

C and D are two point on AB

Also ,AC = CD = DB

As we know that by section formula the co-ordinates of the point which divide internally the line segment joining the point ($x_{1} ,y_{1}$) and ($(x_{2} ,y_{2} )$) in the ratio m:n

So , (x,y) = $(\frac{mx_{2}+nx_{1} }{m+n} ,\frac{my_{2} +ny_{1} }{m+n} )$

Step1:

Given that AC =CD=DB

So, C and D are points of trisection of AB .

Therefore ,D divide the line AB in ratio  2:1.

Now from the formula ,

D = $(\frac{mx_{2}+nx_{1} }{m+n} ,\frac{my_{2} +ny_{1} }{m+n} )$

        =$(\frac{(2)6+(1)3 }{2+1} ,\frac{(2)8 +(1)4}{2+1} )$

        =$(\frac{(12)+3 }{3} ,\frac{(16) +4}{3} ) = (\frac{15}{3} ,\frac{20}{3} )$

D = $(5, \frac{20}{3} )$

Step2:

Now here we can see that ,C divide the line AB in ratio 1:2

So, coordinate of C = $(\frac{mx_{2}+nx_{1} }{m+n} ,\frac{my_{2} +ny_{1} }{m+n} )$

⇒C = $(\frac{(1)6+(2)3 }{1+2} ,\frac{(1)8 +(2)4}{1+2} )$

      = $(\frac{(6+6 }{3} ,\frac{(8 +8}{3} ) = (\frac{12}{3} ,\frac{16}{3} )$

C = $(4, \frac{16}{3})$

Final answer :

Hence , the coordinates of C and D are $(4, \frac{16}{3})$ and $(5, \frac{20}{3} )$.