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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. - Mathematics

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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 

  1. Parallel to the x-axis,
  2. Parallel to the y-axis,
  3. Passing through the origin.
Sum
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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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Chapter 10: Straight Lines - Miscellaneous Exercise [Page 233]

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Chapter 10 Straight Lines
Miscellaneous Exercise | Q 1 | Page 233

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