Find the values of k for which the line (k - 3) x - (4 - K^2) y + k^2 -7k + 6 =0 is

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asked Feb 9, 2018 in Mathematics by Rohit Singh (61,782 points) 35 133 356

Find the values of k for which the line  (k - 3) x - (4 - K2) y + k2 -7k + 6 =0 is

(a) Parallel to the x-axis,
(b) Parallel to the y-axis,
(c) Passing through the origin.

1 Answer

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answered Feb 9, 2018 by sameer (82,980 points) 5 11 37

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

(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-k2)

Now, (k-3) /(4-k2) 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.

Thus, if the given line is passing through the origin, then the value of k is either 1 or 6.

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