Question

please find the answer ​

Answer 1

Answer:

use section formula and take ratio 1:2 i cant see the ques properly so I cant solve

Answer 2

Let the points be A (3,-2) and B (-3,-4) and the points of trisection be P (x,y) and Q (r,s) as shown in the figure.

Now, P divides AB in the ratio 1 : 2

Using the section formula,

$P_{(x,y)} = ( \frac{m x_{2} + n x_{1} }{m + n} , \frac{m y_{2} + n y_{1} }{m + n} )$

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

$= (1, \frac{ - 8}{3} )$

Now point P will be the midpoint of line joining A and Q,

so we can use the midpoint formula

$P(x, y) = \frac{ x_{1} + x_{2}}{2} , \frac{ y_{1} + y_{2} }{2}$

[Here they are coordinates of A and Q]

$= (\frac{3 + r}{2} , \frac{ - 2 + s}{2} )$

But we already know that P(x,y) = P (1,-8/3)

$\frac{3 + r}{2} = 1$

$r = - 1$

Also,

$\frac{ - 2 + s}{2} = \frac{ - 8}{3}$

$s = \frac{ - 10}{3}$

Therefore, the points of trisection are

P(1, -8/3) and Q(-1, -10/3)

Hope this will help, Good luck with your exam!

Have a good evening!