Question

P divides the line segment joining the points a 2, 1 and B 5, - 8 such that AP: AB is equal to 1:3 if P lies on the line 2 -y + K = 0 then find the value of k​

Answer 1

Answer:

The value of k = -4

Step-by-step explanation:

Given: Coordinates of Point A ( 2 , 1 ) and B ( 5 , -8 )

            Ratio of AP : AB = 1 : 3 and equation of a line, 2 - y + k = 0

To find : value of k

To find Value of k  we first find coordinated of point P using section formula.

$Coordinates\:of\:Point P\:=\:(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_2}{m+n})$

where, m : n is ratio in which point divides given points $(x_1,y_1) & (x_2,y_2)$

Given ratio AP : AB = 1 : 3

let AP = 1x and AB = 3x

⇒ AP + BP = 3x

⇒ 1x + BP = 3x

⇒ BP = 3x - 1x

⇒ BP = 2x

⇒ AP : BP = 1 : 2

∴ Coordinates of Point P = $(\frac{5\times1+2\times2}{1+2},\frac{-8\times1+2\times1}{1+2})$

                                         = $(\frac{5+4}{3},\frac{-8+2}{3})$

                                         = $(\frac{9}{3},\frac{-6}{3})$

                                         = $(3,-2)$

Now, using this coordinate and putting value of y in equation , we get

y = -2

⇒ 2 - (-2) + k = 0

⇒ 2 + 2 + k = 0

⇒ 4 + k = 0

⇒ k = -4

Therefore, The value of k = -4