Answer:
Values of k can be 0, 1, or 2 such that intersection of the given lines lie in the 1st quadrant.
Step-by-step explanation:
Given two lines:
[tex]x=2k[/tex] and
[tex]3x+2y=12[/tex]
To find:
Values of 'k' such that the intersection of given two lines lie in the first quadrant.
Solution:
In 1st quadrant, the values of [tex]x[/tex] and [tex]y[/tex] both are positive.
So, let us find out intersection of the two lines.
Intersection of the two lines can be found by solving the two equations for the values of [tex]x[/tex] and [tex]y[/tex].
Given that [tex]x=2k[/tex] to be in the first quadrant, the value of k must be positive.
Let us put [tex]x=2k[/tex] in the equation [tex]3x+2y=12[/tex] to find the intersection point.
[tex]3 \times 2k + 2y=12\\\Rightarrow 6k+2y=12\\\Rightarrow 2y=12-6k\\\Rightarrow \bold{y=6-3k}[/tex]
For y to be positive:
[tex]6 - 3k \geq 0\\\Rightarrow 3k \leq 6\\\Rightarrow k \leq 2[/tex]
So, values of k can be 0, 1, or 2 such that intersection of the given lines lie in the 1st quadrant.
