Question

Consider (4y - X+3) = 0 dividing the line joining points M(-2,5) and N(7, 8) in ratio k: 1. Determine the value of "k"
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Answer 1

Given :- Consider (4y - X+3) = 0 dividing the line joining points M(-2,5) and N(7, 8) in ratio k: 1. Determine the value of "k"

Solution :-

Let (x,y) divides the given line line in k : 1

so, using section formula we get,

→ x = (7k - 2)/(k + 1)

→ y = (8k + 5)/(k + 1)

then, these points must satisfy the line,

→ 4[(8k + 5)/(k + 1)] - [(7k - 2)/(k + 1)] + 3 = 0

→ (32k + 20 - 7k + 2 + 3k + 3)/(k + 1) = 0

→ 28k + 25 = 0

→ 28k = (-25)

→ k = (-25/28) (Ans.)

Hence, value of k will be (-25/28) .

also, since value of k is negative , it will divide the line joining the points externally .

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Answer 2

Given : 4y - X+3 = 0 line dividing the line joining points M(-2,5) and N(7, 8) in ratio k: 1.

To Find : Value of k

Solution:

M(-2,5) and N(7, 8)

Let say point ( a , b)  on line  4y - x+3 = 0   divide the  line joining points M(-2,5) and N(7, 8) in ratio k: 1.

=> a = ( 7k - 2)/(k + 1)     and b = (8k + 5) /(k + 1)

point ( a , b)  on line  4y - x+3 = 0  

=> 4b  - a  + 3  = 0

=> 4  (8k + 5) /(k + 1)   -   ( 7k - 2)/(k + 1)  + 3  = 0

=> 4  (8k + 5)   -  ( 7k - 2)  + 3(k + 1)  = 0

=> 32k + 20 - 7k + 2  + 3k + 3  = 0

=> 28k  + 25  = 0

=>  k  = - 25/28

the value of "k"   is  -25/28

Hence  (4y - X+3) = 0 dividing the line joining points M(-2,5) and N(7, 8) externally

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