Question

Find the coordinates of the points of “trisection” of the line joining the points (6, –2) and (10, 8).​

Answer 1

Answer:

Given coordinates are(x1,y1)=(4,−1)

and (x2,y2)=(−2,−3)

Means of trisection:- A line segment in three equal parts then ratio is 1:2 and 2:1 internally.

Case (1) If m1:m2=1:2

Then, using formula

(x,y)=(m1+m2m1x2+m2x1,m1+m2m1y2+m2y1)

(x,y)=(1+21×(−2)+2×4,1+21×(−3)+2×(−1))

(x,y)=(

Answer 2

Answer:

Trisection coordinates are $(\frac{22}{3},\frac{4}{3}),(\frac{26}{3},\frac{14}{3})$

Step-by-step explanation:

Given coordinates are $(x_1,y_1)=(6,-2)$

and $(x_2,y_2)=(10,8)$

Means of trisection: A line segment in three equal parts then ratio is $1:2$ and $2:1$ internally.

Case $(1)$ If $m_1:m_2=1:2$

Then using formula

$(x,y)=(\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2})$

$(x,y)=(\frac{1(10)+2(6)}{1+2},\frac{1(8)+2(-2)}{1+2})$

$(x,y)=(\frac{10+12}{3},\frac{8-4}{3})$

$(x,y)=(\frac{22}{3},\frac{4}{3})$

Case $(2)$ If $m_1:m_2=2:1$

$(x,y)=(\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2})$

$(x,y)=(\frac{2(10)+1(6)}{1+2},\frac{2(8)+1(-2)}{1+2})$

$(x,y)=(\frac{20+6}{3},\frac{16-2}{3})$

$(x,y)=(\frac{26}{3},\frac{14}{3})$

Trisection coordinates are $(\frac{22}{3},\frac{4}{3}),(\frac{26}{3},\frac{14}{3})$