Question

3.
Let us calculate the ratio in which the point(1,3) divides the line segment joining the points
(4,6) and (3,5)​

Answer 1

Step-by-step explanation:

Let A(4,6) and B(3,5).

Consider a point P(1,3) which is in between A and B. P divides A and B in K:1 ratio.

So,

$( \frac{3k + 4}{k + 1} ) = 1 \\ 3k + 4 = k + 1 \\ 2k = - 3 \\ k = \frac{ - 3}{2}$

So, the ratio is k:1=(-3:2).

Thank you

Answer 2

Given: A line segment formed by joining the points (4, 6) and (3, 5) is divided by the point (1, 3).

To find: The ratio in which it is done so.

Answer:

Let the ratio be k:1.

Section formula:

$\tt \Bigg(\dfrac{mx_2\ +\ nx_1}{m\ +\ n},\ \dfrac{my_2\ +\ ny_1}{m\ +\ n}\Bigg)$

From the given data, we have:

$\tt m\ =\ k\\\\n\ =\ 1\\\\x_1\ =\ 4\\\\x_2\ =\ 3\\\\y_1\ =\ 6\\\\y_2\ =\ 5$

Using them in the formula,

$\tt \Bigg(1,\ 3\Bigg)\ =\ \Bigg(\dfrac{(k\ \times\ 3)\ +\ (1\ \times\ 4)}{k\ +\ 1},\ \dfrac{(k\ \times\ 5)\ +\ (1\ \times\ 6)}{k\ +\ 1}\Bigg)\\\\\\\Bigg(1,\ 3\Bigg)\ =\ \Bigg(\dfrac{3k\ +\ 4}{k\ +\ 1},\ \dfrac{5k\ +\ 6}{k\ +\ 1}\Bigg)$

Equating the x - coordinates, (same can be done with the y - coordinates as well.)

$\tt 1\ =\ \dfrac{3k\ +\ 4}{k\ +\ 1}\\\\\\k\ +\ 1\ =\ 3k\ +\ 4\\\\\\k\ -\ 3k\ =\ 4\ -\ 1\\\\\\-2k\ =\ 3\\\\\\k\ =\ \dfrac{-3}{2}$

Therefore, the line segment formed by joining the points (4, 6) and (3, 5) is divided by the point (1, 3) in the ratio 3:2.