Question

The endpoints of directed line segment PQ have coordinates of P(-7, -5) and Q(5,3). What are the coordinates of point A, on PQ,that divide PQ into a ratio of 1:3? *

Answer 1

Answer:

Therefore the co-ordinate of the point A is (-4,-3)

Step-by-step explanation:

If A(x₁,y₁) and B(x₂,y₂)  is divided by a point in ration m:n internally.

Then the co-ordinate of the point is  $(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_2}{m+n})$.

Given two points are P(-7,-5) and Q(5,3).

A point A on PQ, that divides PQ into a ratio of 1:3.

Since the point A lies on the PQ.

Therefore it divides the line segment PQ internally.

Here x₁= -7,y₁= -5 ,  x₂= 5,y₂ =3, m=1,n=3

Therefore the co-ordinate of the point A is

$(\frac{5.1+(-7).3}{1+3},\frac{3.1+(-5).3}{1+3})$

$=(\frac{-16}{4},\frac{-12}{4})$

=(-4,-3)

Answer 2

The coordinates of point A on PQ that divides the line PQ in the ratio of 1:3 is (-4,-3)

The given information are:

Coordinates of P(-7, -5)

Coordinates of Q(5,3)

Where,

$x_{1}=-7\\x_{2}=5\\y_{1}=-5\\y_{2}=3$

The formula to calculate the coordinates of point A is:

$\text{Coordinates of A} = (\dfrac{ mx_{2} +nx_{1}}{m+n}:\dfrac{my_{2}+ny_{1}}{m+n})$

$\text{Coordinates of A} = (\dfrac{ 1\times5_{} +3\times-7\_{}}{1+3}:\dfrac{1\times3_{}+3\times-5_{}}{1+3})$

$\text{Coordinates of A} = (\dfrac{5+(-21)}{4}:\dfrac{3-15}{4})$

$\text{Coordinates of A} = (\dfrac{-16}{4}:\dfrac{-12}{4})$

$\text{Coordinates of A} = (-4:-3})$

Therefore, the coordinates of point A are (-4,-3).

To know more about the calculation of the coordinates, refer to the link below:

https://brainly.com/question/23689760