Question

Can someone help me please?

Answer 1

Given:

A line segment AB.

A point is $\dfrac{3}{10}$ of the way from A to B.

To find:

The coordinates of that point.

Solution:

Section formula: If a point divide a line segment in m:n, then

$Point=\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)$

From the given figure, it is clear that the two end points of the line segment are A(-4,-5) and B(12,4).

Let the unknown point be P. The point is $\dfrac{3}{10}$ of the way from A to B. It means,

$\dfrac{AP}{AB}=\dfrac{3}{10}$

Let AP and AB are 3x and 10x, then

$\dfrac{AP}{PB}=\dfrac{AP}{AB-AP}$

$\dfrac{AP}{PB}=\dfrac{3x}{10x-3x}$

$\dfrac{AP}{PB}=\dfrac{3x}{7x}$

$\dfrac{AP}{PB}=\dfrac{3}{7}$

It means, point P divided the line segment in 3:7.

Using section formula, we get

$P=\left(\dfrac{3(12)+7(-4)}{3+7},\dfrac{3(4)+7(-5)}{3+7}\right)$

$P=\left(\dfrac{36-28}{10},\dfrac{12-35}{10}\right)$

$P=\left(\dfrac{8}{10},\dfrac{-23}{10}\right)$

$P=\left(0.8,-2.3\right)$

Therefore, the coordinates of the required point are (0.8, -2.3).