Question

find the points of trisection of the line joining A (2,-3) B(4,5)​

Answer 1

$(\frac {8}{3}, \frac {-1}{3})\ and\ (\frac {10}{3},\frac {7}{2})$ are the Points

Step-by-step explanation:

Given,

A(2,-3)and B(4,5) are the points of the line segment

Let us suppose that the points of trisection of AB are P & Q

i.e., AP = PQ = QB

Hence, AB is divided internally in the ratio 1 : 2 through P.

Therefore, the coordinates of P, by applying the section formula, are

$(\frac {mx_2 + nx_1}{m+n},\frac {my_2+ny_1}{m+n})$

= [$\frac {1(4)+2(2)}{1+2}$], [$\frac {1(5)+2(-3)}{1+2}$]

= $[\frac {4+4}{3}],[\frac {5-6}{3}]$

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

Now, Q also divides AB internally in the ratio 2 : 1. So, the coordinates of Q are

= $[\frac {2(4)+1(2)}{1+2}, \frac {2(5)+1(-3)}{1+2}]$

= $[\frac {10}{3},\frac {7}{2}]$

Therefore, the coordinates of the points of trisection of the line segment joining A and B are

$(\frac {8}{3}, \frac {-1}{3}) and (\frac {10}{3},\frac {7}{2})$

Answer 2

Step-by-step explanation:

hope it helps you!

thankyou.