For what value of k (k > 0) will the lines 5x + ky - 2 = 0 and (k + 1)x + 4y + 7 = 0 be parallel?

Correct option is A

Given:
5x + ky - 2 = 0
(k + 1)x + 4y + 7 = 0
k > 0
Formula Used:
In line equation, y = mx + c, where m is slope.
Slope of line 1 = Slope of line 2 for parallel condition.
Solution:
For first line, 5x + ky - 2 = 0
​$$y = \frac{-5x}{k}+\frac{2}{k}$$​​
Slope of line, $$m_1= \frac{-5}{k}$$​​
For second line, (k + 1)x + 4y + 7 = 0
​$$y = \frac{-(k+1)}{4}-\frac{7}{4}$$​​
Slope of line, $$m_2=\frac{-(k+1)}{4}$$​​
For parallel condition, $$m_1=m_2$$​​
​$$\frac{-5}{k}=-\frac{(k+1)}{4} \\\ \\5×4=k(k+1) \\\ \\20=k^2+k \\\ \\k^2+k-20=0$$​​
Splitting the middle term, we get
​$$k^2+5k-4k-20=0 \\\ \\k(k+5)-4(k+5)=0 \\\ \\(k+5)(k-4)=0 \\\ \\K = 4 or -5$$​​
As per the question k>0, so k = 4.