Question

Find the values of k for which the line $\left( k-3 \right)x-\left( 4-{{k}^{2}} \right)y+{{k}^{2}}-7k+6=0$
$\begin{align}
  & \text{a) parallel to x-axis} \\
 & \text{b) parallel to y-axis} \\
 & \text{c) Passing through origin}\text{.} \\
\end{align}$

Answer 1

Hint: Now we are given with a linear equation in two variables. We know that for a line ax + by + c = 0, the line is parallel to the x-axis if a = 0, the line is parallel to the y-axis if b = 0 and the line passes through origin if c = 0. Hence we will use this condition to find k.

Complete step by step answer:
Now consider any general linear equation in two variables ax + by + c = 0.
Now we know that if the line is parallel to x axis then a = 0. If the line is parallel to y-axis then b = 0. And if the line passes through origin c = 0.
Now Let us consider the given line.
We have $\left( k-3 \right)x-\left( 4-{{k}^{2}} \right)y+{{k}^{2}}-7k+6=0$
Now if we compare the line with ax + by + c we have.
$a=\left( k-3 \right)$ , $b=\left( 4-{{k}^{2}} \right)$ and $c={{k}^{2}}-7k+6$
Now if the line is parallel to x-axis we have a = 0.
Hence using this condition we have
$k-3=0\Rightarrow k=3$
Hence the lines is parallel to the x-axis if k = 3.
Now similarly the line is parallel to y-axis if b = 0.
Hence we have
$\begin{align}
  & 4-{{k}^{2}}=0 \\
 & \Rightarrow {{k}^{2}}=4 \\
 & \therefore k=\pm 2 \\
\end{align}$
Hence we have the line is parallel to y-axis if $k=\pm 2$
Now for line to pass through origin we have c = 0.
Hence we get.
$\begin{align}
  & {{k}^{2}}-7k+6=0 \\
 & \Rightarrow {{k}^{2}}-6k-k+6=0 \\
 & \Rightarrow k\left( k-6 \right)-1\left( k-6 \right)=0 \\
 & \Rightarrow \left( k-1 \right)\left( k-6 \right)=0 \\
\end{align}$
Hence k = 6 or k = 1.

Hence we have the line passes through origin for k = 6 and k = 1.

Note: Now the conditions that we have for lines to be parallel to x, y axis and passing through origin can easily be understood
Now the line which is parallel to x-axis is of the form y = c.
Hence we must have a coefficient of x as 0. Similarly the line parallel to y axis must have a coefficient of y to be 0.
Now the line which passes through origin will satisfy (0, 0). Hence if we substitute (0, 0) in ax + by + c = 0 we get c = 0.