Question

find the co ordinates of the points of trisection of the line segment joining the points (-2,1) and (7,4)​

Answer 1

Concept :-

Here the concept of Trisection points have been used . Trisection points means The points which divides the line segment in ratio 1 : 2 and 2 :1 Let A= (-2 ,1) and B=(7, 4) Lets find trisection points cordinates using section formula One time we shall take ratio 1 : 2 and other time 2 : 1 Then we can get co-ordinates of trisection points Let the of trisection points be P,Q

To find :-

Co-ordinates of trisection points

Section formula:-

$\dfrac{mx_2+nx_1}{m+n}$, $\dfrac{my_2+ny_1}{m+n}$

Solution:-

${P}$ = $\dfrac{mx_2+nx_1}{m+n}$, $\dfrac{my_2+ny_1}{m+n}$

${x_1= -2}$

${x_2 = 7}$

${y_1 = 1}$

${y_2 =4}$

m = 1

n =2

${P}$ = $\dfrac{1(7)+2(-2)}{1+2}$, $\sf\dfrac{1(1)+2(4)}{1+2}$

${P}$ = $\dfrac{7-4}{3}$, $\sf\dfrac{1+8}{3}$

${P}$=$\dfrac{3}{3}$, $\sf\dfrac{9}{3}$

P = (1 , 3)

${Q}$ = $\dfrac{mx_2+nx_1}{m+n}$, $\dfrac{my_2+ny_1}{m+n}$

${x_1= -2}$

${x_2 = 7}$

${y_1 = 1}$

${y_2 =4}$

m = 2

n = 1

${Q}$ = $\dfrac{2(7) +1(-2)}{2+1}$, $\sf\dfrac{2(4)+1(1)}{3}$

${Q}$ = $\dfrac{14-2}{3}$, $\sf\dfrac{8+1}{3}$

${Q}$ = $\dfrac{12}{3}$, $\sf\dfrac{9}{3}$

Q = (4 , 3)

So, the co-ordinates of trisection points P, Q = (1 ,3) and (4 ,3)