Question

if the origin cuts the line segment points p (-2,-4) and q(6,k) in the ratio 1:3 find the value of k.​

Answer 1

Answer:

The value of k = 4/3

Step-by-step explanation:

Internal section formula:

   Let A(x₁,y₁) and B(x₂,y₂) be two points of a line segment and C(x,y) is a point which divides the line segment internally in the ratio m:n then the Internal section formula is given by

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

   Given,

    Let O(0,0) be the origin.

    Line segment joining the points p(-2,-4) and q(6,k).

  the origin O(0,0) divides the line internally in the ratio 1:3.

     By the formula we can write

            $(0,0) = (\frac{1(6)+3(-2)}{1+3} ,\frac{3(k)+1(-4)}{1+3} )$

            (0,0) = ((6-6/2),(3k-4)/4)

        By equating on both sides

                     (3k-4)/4 = 0

                          3k-4 = 0

                             3k = 4

                               k = 4/3

          Hence, the value of k = 4/3

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Answer 2

Answer:

given:-

p(-2,-4)=(x1,y1)

q(6,k)=(x2,y2)

m=1

n=3

(x,y)=(0,0) point of intersection

we know that :-

(x,y) = (mx2+nx1/m+n,my2+ny1/m+n)

therefore

(0,0)=(1×6+3×-2/1+3,1×k+3×-4/1+3)

(0,0)=(6-6/4,k-12/4)

(0,0)=(0,k-12/4)

by equating

0=k-12/4

0=k-12

k=12.