Question

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

Answer 1

The given equation of line is 
\((k-3) x - (4 - k^ 2 ) y + k ^2- 7k + 6 = 0 … (1)\)

(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 
\((4- k^ 2 ) y = (k- 3) x + k^ 2 - 7k + 6 = 0\)

\(y=\frac{(k-3)}{(4-k^2)}x+\frac{k^2-7k+6}{(4-k^2)}\)

∴ Slope of the given line =\(\frac{(k-3)}{(4-k^2)},\) which is of the form \(y = mx + c. \)
Slope of the x-axis = 0 

\(∴ \frac{(k-3)}{(4-k^2)}=0\)

\(⇒ k-3=0\)

\(⇒ k=3\)
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  \(\frac{(k-3)}{(4-k^2)}\)

Now, \(\frac{(k-3)}{(4-k^2)}\) is undefined at  \(k^2 = 4 \)

\(k^2 = 4 \)
\(⇒ k = ±2 \)
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.

\((k-3)(0)-(4-k^2)(0)+k^2-7k+6=0\)

\(k^2-7k+6=0\)

\((k-6)(k-1)=0\)

\(k=1\space  or\space 6\)

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