Question

pls answer fast


I'll mark as brainliest................​

Answer 1

Answer:

( 23 / 3, 20 / 3 ) is one trisection point (closer to (11, 9))

( 13 / 3, 13 / 3 ) is the other trisection point (closer to (1, 2))

Step-by-step explanation:

( ( 2×11 + 1 ) / 3,  ( 2× 9 + 2 ) / 3 )

= ( 23 / 3,  20 / 3 )

( ( 11 + 2×1 ) / 3, ( 9 + 2×2 ) / 3 )

= ( 13 / 3,  13 / 3 )

Answer 2

       

Let the points be (a, b) and (m, n).

Let (a, b) be after (1, 2), and (m, n) be before (11, 9).

Let  P = (1, 2),   Q = (a, b),   R = (m, n),   and S = (11, 9).

We know that the coordinates of collinear points equidistant from each other are in AP.

So, as Q and R trisect the line segment, PQ = QR = RS. Therefore, x coordinates of P, Q, R and S are in AP. So are y coordinates.

Consider the AP  1, a, m, 11,..., formed by x coordinates.

$x_1=1 \\ \\ x_2=a \\ \\ x_3=m \\ \\ x_4=11$

$x_4-x_1 =11-1 \\ \\ (4-1)d=10 \\ \\ 3d=10 \\ \\ d=\frac{10}{3} \\ \\ d=3\frac{1}{3} \\ \\ \\ x_2=x_1+d \\ \\ a=1+3\frac{1}{3} \\ \\ a=\bold{4\frac{1}{3}} \\ \\ \\ x_3=x_4-d \\ \\ m=11-3\frac{1}{3} \\ \\ m=\bold{7\frac{2}{3}}$

Consider the AP 2, b, n, 9,..., formed by y coordinates.

$y_1=2 \\ \\ y_2=b \\ \\ y_3=n \\ \\ y_4=9$

$y_4-y_1=9-2 \\ \\ (4-1)d=7 \\ \\ 3d=7 \\ \\ d=\frac{7}{3} \\ \\ d=2\frac{1}{3} \\ \\ \\ y_2=y_1+d \\ \\ b=2+2\frac{1}{3} \\ \\ b=\bold{4\frac{1}{3}} \\ \\ \\ y_3=y_4-d \\ \\ n=9-2\frac{1}{3} \\ \\ n=\bold{6\frac{2}{3}}$

∴ Q = (a, b) =  $\bold{(4\frac{1}{3},\ 4\frac{1}{3})}$

&  R = (m, n) =  $\bold{(7\frac{2}{3},\ 6\frac{2}{3}) }$

Hope this helps. Plz ask me if you've any doubts.

Thank you. :-))