Question

Find the coordinates of the points of trisection of the line segment (4, – 1) and (– 2, 3).

Answer 1

Let the points be A(4,–1) and B(–2,3)

PQ are on AB such that

AP = PQ = QB = m

Point P divides AP & PB in the ratio

AP = m

PB = PQ + QB

= k + k

= 2k

Hence,

Ratio between AP & PB = AP/PB = 1/2

Thus P divides AB in the ratio 1:2

Finding P

Let P(x,y)

$\rm \: m_{1} \: = 1 \: an d \: \: m_{2} = 2$

And for AB

$\rm \: x_{1} = 4 \: \: and \: \: x_{2} = - 2$

$\rm\: y_{1} = - 1 \: \: and \: \: y_{2} = -3$

$\rm \: x = \frac{m_{1}x_{2} + {m_{2}x_{1}}}{m_{1} + m_{2}}$

$x = 2$

$\rm \: y = \frac{m_{1}y_{2} + {m_{2}y_{1}}}{m_{1} + m_{2}}$

$y = \frac{ - 5}{3}$

Similarly Q divides AQ and QB

$\rm \: \frac{AQ}{QB} = \frac{AP+PQ}{QB}$

$= \frac{k + k}{k}$

$\frac{2k}{k}$

$2 : 1$

Finding Q

Let Q be (x,y)

$\rm \: m_{1} \: = 2 \: an d \: \: m_{2} = 1$

$\rm \: x_{1} = 4 \: \: and \: \: x_{2} = - 2$

$\rm \: y_{1} = - 1 \: \: and \: \: y_{2} = - 3$

$\rm \: x = \frac{m_{1}x_{2} + {m_{2}x_{1}}}{m_{1} + m_{2}}$

$x = 0$

$\rm \: y = \frac{m_{1}y_{2} + {m_{2}y_{1}}}{m_{1} + m_{2}}$

$y = \frac{ - 7}{3}$

Hence point Q is Q(x,y)

$Q = 0 \: and \: \frac{ - 7}{3}$

Answer 2

Answer:

hope it will help you

Step-by-step explanation:

Find the coordinates of the points of trisection of the line segment (4, – 1) and (– 2, 3).