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Question: Find the values of k for which the line \(\left( k-3 \right)x-\left( 4-{{k}^{2}} \right)y+{{k}^{2}}-...

Find the values of k for which the line (k−3)x−(4−k2)y+k2−7k+6=0\left( k-3 \right)x-\left( 4-{{k}^{2}} \right)y+{{k}^{2}}-7k+6=0
a) parallel to x-axis b) parallel to y-axis c) Passing through origin. \begin{aligned} & \text{a) parallel to x-axis} \\\ & \text{b) parallel to y-axis} \\\ & \text{c) Passing through origin}\text{.} \\\ \end{aligned}

Explanation

Solution

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 (k−3)x−(4−k2)y+k2−7k+6=0\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=(k−3)a=\left( k-3 \right) , b=(4−k2)b=\left( 4-{{k}^{2}} \right) and c=k2−7k+6c={{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⇒k=3k-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
4−k2=0 ⇒k2=4 ∴k=±2 \begin{aligned} & 4-{{k}^{2}}=0 \\\ & \Rightarrow {{k}^{2}}=4 \\\ & \therefore k=\pm 2 \\\ \end{aligned}
Hence we have the line is parallel to y-axis if k=±2k=\pm 2
Now for line to pass through origin we have c = 0.
Hence we get.
k2−7k+6=0 ⇒k2−6k−k+6=0 ⇒k(k−6)−1(k−6)=0 ⇒(k−1)(k−6)=0 \begin{aligned} & {{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{aligned}
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.