Question

1. Given a triangle ABC in which A = (4. - 4),
B = (0, 5) and C = (5, 10). A point P lies on
BC such that BP: PC = 3:2. Find the length
of line segment AP​

Answer 1

Answer:

√145

Step-by-step explanation:

Given,

A = (4 , - 4)

B = (0 , 5)

C = (5 , 10)

BP : PC = 3 : 2

To Find :-

Length of line segment AP.

How To Do :-

We need to find the co - ordinates of 'P' by using Internal Division formula(Section formula). After obtaining the co-ordinates of 'P' we need to find the length of AP using distance formula.

Formula Required :-

Internal Division Formula :-

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

Distance Formula :-

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Solution :-

BP : PC = 3 : 2

→ m : n = 3 : 2

B = (0 , 5)

Let,

x₁ = 0 , y₁ = 0

C = (5 , 10)

Let,

x₂ = 5 , y₂ = 10

Substituting in Internal Division Formula :-

$P=\left(\dfrac{3(5)+2(0)}{3+2},\dfrac{3(10)+2(5)}{3+2}\right)$

$=\left(\dfrac{15+0}{5},\dfrac{30+10}{5}\right)$

$=\left(\dfrac{15}{5},\dfrac{40}{5}\right)$

∴ P = (3 , 8)

AP :-

A = (4 , -4)

Let,

x₁ = 4 , y₁ = - 4

P = (3 , 8)

Let,

x₂ = 3 , y₂ = 8

Substituting values in Distance formula :-

$AP=\sqrt{(3-4)^2+(8-(-4))^2}$

$=\sqrt{(-1)^2+(12)^2}$

$=\sqrt{1+144}$

= √145

Length of AP = √145