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Question
Find the values of k for which the line (k–3) x – (4 – k2) y + k2 –7k + 6 = 0 is
- Parallel to the x-axis,
- Parallel to the y-axis,
- Passing through the origin.
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Solution
The given equation of line is
(k – 3) x – (4 – k2) y + k2 – 7k + 6 = 0 …(1)
(a) If the given line is parallel to the x-axis, then
Slope of the given line = Slope of the x-axis
The given line can be written as
(4 – k2) y = (k – 3) x + k2 – 7k + 6 = 0
`y = ((k - 3))/((4 - k^2))x + (k^2 - 7k + 6)/((4 - k^2))`, which of the form y = mx + c
∴ Slope of the given line = `(k - 3)/(4 - k^2)`
Slope of the x-axis = 0
∴ `((k - 3))/((4 - k^2)) = 0`
= k - 3 = 0
= k = 3
Thus, if the given line is parallel to the x-axis, then the value of k is 3.
(b) If the given line is parallel to the y-axis, it is vertical. Hence, its slope will be undefined.
The slope of the given line is `(k - 3)/(4 - k^2)`.
Now, `(k - 3)/(4 - k^2)` is undefined at k2 = 4
k2 = 4
⇒ k = ±2
Thus, if the given line is parallel to the y-axis, then the value of k is ±2.
(c) If the given line is passing through the origin, then point (0, 0) satisfies the given equation of line.
(k - 3) (0) - (4 - k2)(0) + k2 - 7k + 6 = 0
k2 - 7k + 6 = 0
k2 - 6k + k + 6 = 0
(k - 6) (k - 1) = 0
k = 1 or 6
Thus, if the given line is passing through the origin, then the value of k is either 1 or 6.
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