Question

2. A (-3, 4), B (3, -1) and C (-2, 4) are the vertices of a triangle ABC.
Find the length of line segment AP, where point P lies inside
BC, such that BP: PC = 2:3.
.
.

plz answer n show steps .....ll b a great help​

Answer 1

Solution :-

Draw the given ΔABC with the given vertices

[ Refer to attachment ]

To find the length of line segment AP we need both the coordinates of the line segment

Coordinates of 'P' are not given.

So, we have to find the coordinates of 'P' with the given information

'P' is a points on BC which divides BC is the ratio of 2 : 3

B(3,-1) C(-2,4) are given, we can find the coordinates of P using

Section formula :

$P(x,y) = \bigg( \dfrac{mx_2 + nx_1 }{m + n} ,\dfrac{my_2 + ny_1 }{m + n} \bigg)$

Here, x₁ = 3, x₂ = - 2, y₁ = - 1, y₂ = 4

m : n = 2 : 3 ⇒ m = 2, n = 3

$\implies P(x,y) = \bigg( \dfrac{2( - 2) + 3(3) }{2 + 3} ,\dfrac{2(4) + 3( - 1) }{2 + 3} \bigg)$

$\implies P(x,y) = \bigg( \dfrac{ -4 + 9}{5} ,\dfrac{8 - 3 }{5} \bigg)$

$\implies P(x,y) = \bigg( \dfrac{5}{5} ,\dfrac{5 }{5} \bigg) = (1,1)$

Now A(- 3, 4) and P(1,1) are known

Now we can find the length of AP using

Distance formula :

$d = \sqrt{( x_2 - x_1)^{2} + (y_2 - y _1) ^{2} }$

Here, x₁ = - 3, x₂ = 1, y₁ = 4, y₂ = 1

$\implies AP = \sqrt{ \{ 1-( - 3) \}^{2} + (1 - 4) ^{2} }$

$\implies AP = \sqrt{(1 + 3) ^{2} + ( - 3) ^{2} }$

$\implies AP = \sqrt{4^{2} + 9 }$

$\implies AP = \sqrt{16 + 9 }$

$\implies AP = \sqrt{25} = 5$

Therefore the length of AP is 5 units.