Question

if the line (2x+3y+5)+k(x-7y+6)=0 is parallel to xaxis then k=​

Answer 1

Answer:

$\large\boxed{\sf{k=-2}}$

Step-by-step explanation:

Given an equation of line such that,

$(2x + 3y + 5) + k(x - 7y + 6) = 0$

On simplifying, we will get,

$= > (2 + k)x + (3 - 7k)y + (5 + 6k) = 0$

Now, this line is parallel to X axis.

Therefore, it's slope must be equal to 0.

Differentiating the eqn wrt x, we get,

$= > ( 2 + k) + (3 - 7k) \dfrac{dy}{dx} = 0 \\ \\ = > \dfrac{dy}{dx} = - \dfrac{(2 + k)}{(3 - 7k)} \\ \\ = > \dfrac{dy}{dx} = \dfrac{2 + k}{7k - 3}$

Thus, we will equate this slope to zero.

Therefore, we will get,

$= > \dfrac{2 + k}{7k - 3} = 0 \\ \\ = > 2 + k = 0 \\ \\ = > k = - 2$

Hence, required value of k = -2

Answer 2

Answer:

hope you find the attachment