Question

Let P and Q be the points of trisection of the line segment joining the points A(2, – 2) and B(–7, 4) such that P is nearer to A. The coordinates of P and Q are given by​

Answer 1

Since P and Q be the points of trisection of the line segment joining the points A(2,−2) and B(−7,4) such that P is nearer to A.

Therefore, P divides line segment in the ratio 1:2 and Q divides in 2:1 as shown in the figure.

Using section formula,

Coordinates of P = [ m+nmx 2 +nx 1, m+nmy 2+ny 1 ]= [ 1+21(−7)+2(2) , 1+21(4)+2(−2) ]= [ 3−3 , 30 ]= [−1,0]

coordinates of Q =[ 2+12(−7)+1(2), 2+12(4)−1(−2) ]

= [ 3−12 , 36 ]

= [−4,2]

its help for you

Answer 2

Step-by-step explanation:

according to question

AP=PQ=QB

• hence P divide AB 1:2ratio
• now use section formula

p (x, y)

x=(1 x (-7)+2 x 2)/(2+1)=-3/3=-1

y=(1 x4+2 x (-2)) /(2+1)=0

p (-1,0)

now Q divide line AB 2:1ratio

use again section formula

Q (r, u)

r=(2 x (-7)+1 x 2)/(2+1)=-12/3=-4

u=(2 x4+1 x (-2))/(2+1)=2

Q (-4,2)