NCERT Solutions
Class 11 Maths
Straight Lines

Ex.Misc. Q.1
Find the values of k for which the line is (k - 3) x – (4 – k2) y + k2 – 7k + 6 = 0
(a) Parallel to the x-axis,
(b) Parallel to the y-axis,
(c) Passing through the origin.
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, which is of the form y = mx + c.
Slope of the given line = (k - 3) ÷ (4 – k2)
Slope of the x-axis = 0
So, (k - 3) ÷ (4 – k2) = 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 – 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.
(k - 3) × 0 – (4 – k2) × 0 + k2 – 7k + 6 = 0
=> k2 – 7k + 6 = 0
=> (k - 6) (k - 1) = 0
=> k = 6, 1
Thus, if the given line is passing through the origin, then the value of k is either 1 or 6.