Question

Let A(2, 2), B(8, –4) be two given points in a plane. If a point P lies on the X- axis (in

positive side), and divides AB in the ratio 1: 2, then find the coordinates of P.​

Answer 1

Answer:

(6,-2)

Step-by-step explanation:

X=mx2+nx1/m+n.and y=my2+ny1/m+n

2×8+1×2/2+1. , 2×-4+1×2/2+1

16+2/3, - 8+2/3

18/3, - 6/3

6,-2

Answer 2

Answer:

The coordinates of P are (4, 0).

Step-by-step explanation:

Given:

Coordinates of A = (2, 2)

Coordinates of B = (8, -4)

m:n = 1:2

To find:

Coordinates of P

Formula:

Internal section division

$P(x, y) =( \frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n} )$

Solution:

Let $(x_1, y_1)$ = (2, 2) and $(x_2,y_2)$ = (8, -4).

Also, m = 1 and n = 2

Substituting the given values in the above formula, we get

$P(x, y) =( \frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n} )$

$P(x, y) =( \frac{(1)(8)+(2)(2)}{1+2}, \frac{(1)(-4)+(2)(2)}{1+2} )$

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

$P(x, y) =( \frac{12}{3}, \frac{0}{3} )$

$P(x, y) =( 4, 0 )$

Therefore, point P lies at (4, 0) on the X-axis.

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