Question

Find the Value of k so that lines kx+2y=1 and 3x+k²y=-2 are perpendicular (k≠0)​

Answer 1

Answer: k=-3/2

Step-by-step explanation:

(1) kx+2y=1 => y=(-k/2)x+1/2

(2) 3x+(k^2)y=-1 => y=(-3/k^2)x-1/(k^2)

(1) and (2) are perpendicular if (-k/2)×(-3/k^2)=-1 => k=-3/2

Answer 2

Answer:

The Value of k so that lines kx+2y=1 and 3x+k²y=-1 are perpendicular is $\dfrac{-3}{2}$

Step-by-step explanation:

Given equations,

• kx + 2y=1 ------------(i)
• $3x + k^{2}y = -1$ ---------------------(ii)

We have to find the value of x:-

In equation (i)-

kx + 2y=1

2y = 1 - kx

$y = \dfrac{-kx+1}{2}$

In equation (ii)-

$3x + k^{2}y = -1$

$k^{2}y = -1 - 3x$

$y = \dfrac{-3x-1}{k^2}$

Both equation is perpendicular to each other.

Therefore, the coefficient of x in eq-(i)$\times$ the coefficient of x in eq-(ii) = -1

$\dfrac{-k}{2}\times\dfrac{-3}{k^2} = -1$

$\dfrac{-1}{2}\times\dfrac{-3}{k} = -1$

3 = -2k

k = $\dfrac{-3}{2}$