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. Find the the coordinates of P and Q

Answer 1

Given:- P and Q are points of trisection of the line segment joining the points A(2,-2) and B(-7,4) in such a way that P is nearer to A.

Solution:-

Since P and Q are the points of trisection so we cy say that point P divides the line segment in the ratio of 1:2 whereas Q divides the line segment in the ratio of 2:1.

Now, apply section formula here,

$\frac{mx2 + nx1}{m + n} \: \frac{my2 + ny1}{m + n}$

Coordinates of P we have,

= [mx2+nx1/m+n , my2+ny1/m+n]

= [1×(-7)+2×(2)/1+2, 1×(4)+2×(-2)/1+2]

= [-3/3 , 0/3]

= [-1, 0]

And now for the coordinates of Q we have,

= [mx2+nx1/m+n , my2+ny1/m+n]

=[2×(-7)+ 1× (2)/ 2+1 , 2×(4) - 1×(-2)/2+1]

=[-12/3 , 6/3 ]

=[-4,2]

Therefore, coordinates of P [-1, 0] and the coordinates of Q [-4,2].

Answer 2

Answer:

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)