Question

The ratio in which p (4,m) divides the line segment joining the point A (2,3) and B (6,-3). Hence find the value of m

Answer 1

Given: Point P(4, m) dividing the points A(2, 3) and B(6, -3).

To find: The ratio in which it is divided and the value of m.

Answer:

Section formula:

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

Let's assume the ratio is k:1. Now, from the given data, we have:

$\tt m\ =\ k\\\\n\ =\ 1\\\\x_1\ =\ 2\\\\y_1\ =\ 3\\\\x_2\ =\ 6\\\\y_2\ =\ -3$

Using them in the section formula,

$\tt P(4, m)\ =\ \bigg(\dfrac{(k\ \times\ 6)\ +\ (1\ \times\ 2)}{k\ +\ 1},\ \dfrac{(k\ \times\ -3)\ +\ (1\ \times\ 3)}{k\ +\ 1}\bigg)\\\\\\\\P(4, m)\ =\ \bigg(\dfrac{6k\ +\ 2}{k\ +\ 1},\ \dfrac{-3k\ +\ 3}{k\ +\ 1}\bigg)$

Equating the x coordinates,

$\tt 4\ =\ \dfrac{6k\ +\ 2}{k\ +\ 1}\\\\\\4k\ +\ 4\ =\ 6k\ +\ 2\\\\\\4\ -\ 2\ =\ 6k\ -\ 4k\\\\\\2\ =\ 2k\\\\\\k\ =\ 1$

Therefore, the ratio is 1:1.

Now that we have k, we can find the value of m.

Equating the y coordinates,

$\tt m\ =\ \dfrac{-3k\ +\ 3}{k\ +\ 1}\\\\\\m\ =\ \dfrac{-3\ +\ 3}{2}\\\\\\2m\ =\ -3\ +\ 3\\\\\\m\ =\ 0$

Therefore, the ratio is 1:1 and the value of m is 0.