If the equations 4x + (k − 2)y + 3 = 0 and (k − 2)x + 9y − 5 = 0 (k 0) are parallel, then find the value of k^2 + 6.

Question

If the equations 4x + (k − 2)y + 3 = 0 and (k − 2)x + 9y − 5 = 0 (k > 0) are parallel, then find the value of k^2 + 6.

Accepted Answer

Given two lines:
1) 4x + (k - 2)y + 3 = 0
2) (k - 2)x + 9y - 5 = 0

For two lines to be parallel, their slopes must be equal.

Slope of line 1:
Rewrite as y = mx + c form:
4x + (k - 2)y + 3 = 0
=> (k - 2)y = -4x - 3
=> y = (-4/(k - 2))x - 3/(k - 2)
Slope m1 = -4/(k - 2)

Slope of line 2:
(k - 2)x + 9y - 5 = 0
=> 9y = -(k - 2)x + 5
=> y = (-(k - 2)/9)x + 5/9
Slope m2 = -(k - 2)/9

Set slopes equal for parallelism:
-4/(k - 2) = -(k - 2)/9
Multiply both sides by 9(k - 2):
-4 * 9 = -(k - 2)^2
-36 = -(k - 2)^2
Multiply both sides by -1:
36 = (k - 2)^2

Take square root:
k - 2 = ±6

Since k > 0:
Case 1: k - 2 = 6 => k = 8
Case 2: k - 2 = -6 => k = -4 (not valid as k > 0)

So, k = 8

Calculate k^2 + 6:
= 8^2 + 6
= 64 + 6
= 70

Hence, the correct answer is 70.

Answer

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