Question

Given points A(-7, 4) and B(5, 12), find the point P that partitions directed line segment top enclose A B end enclose in the ratio of 3:1. Type your answer as an ordered pair (x,y).

Answer 1

Answer:

Therefore,

$P(x,y)=(2,10)$

Step-by-step explanation:

Given:

Let Point P ( x , y ) divides Segment AB in the ratio 3 : 1 = m : n (say)

point A( x₁ , y₁) ≡ ( -7 , 4)  

point B( x₂ , y₂) ≡ (5 , 12)  

To Find:  

point P( x , y) ≡ ?  

Solution:  

IF a Point P divides Segment AB internally in the ratio m : n, then the Coordinates of Point P is given by Section Formula as

$x=\dfrac{(mx_{2} +nx_{1}) }{(m+n)}\\ \\and\\\\y=\dfrac{(my_{2} +ny_{1}) }{(m+n)}$

Substituting the values we get

$x=\dfrac{(3\times 5 +1\times -7) }{(3+1)} \ \ \ and\ \ \ y=\dfrac{(3\times 12 +1\times 4) }{(3+1)}\\\\\\\therefore x = \dfrac{8}{4}=2 \ \ and\ \ \therefore y = \dfrac{40}{4}=10\\\\\\\therefore P(x,y) = (2 , 10)$

Therefore,

$P(x,y)=(2,10)$