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(a) Parallel to the $x-axis$ .

(b) Parallel to the $y-axis$ .

(c) Passing through the origin.

Answer

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Hint: For solving this problem we will directly apply 3 concepts from the co-ordinate geometry and then we will solve for the value of $k$ for each case.

Complete step-by-step solution -

Given:

There is an equation of the straight line $\left( k-3 \right)x-\left( 4-{{k}^{2}} \right)y+{{k}^{2}}-7k+6=0$ . We have to find the value of $k$ for given cases. Before we start solving first look at the following points about a general straight line $L:ax+by+c=0$ :

1. If $L$ passes through a point $\left( h,k \right)$ then, $ah+bk+c=0$ .

2. If $L$ is parallel to the $x-axis$ then, the coefficient of $x$ will be 0 $\left( a=0 \right)$ .

3. If $L$ is parallel to the $y-axis$ then, the coefficient of $y$ will be 0 $\left( b=0 \right)$ .

Now, we will solve each part one by one using the above mentioned points. Then,

(a) Parallel to the $x-axis$ :

If $\left( k-3 \right)x-\left( 4-{{k}^{2}} \right)y+{{k}^{2}}-7k+6=0$ is parallel to the $x-axis$ . Then, from the second point, we can say that the coefficient of $x$ will be 0. Then,

$\begin{align}

& k-3=0 \\

& \Rightarrow k=3 \\

\end{align}$

Thus, when $k=3$ then the given line will be parallel to $x-axis$ .

(b) Parallel to the $y-axis$ :

If $\left( k-3 \right)x-\left( 4-{{k}^{2}} \right)y+{{k}^{2}}-7k+6=0$ is parallel to the $y-axis$ . Then, from the third point, we can say that the coefficient of $y$ will be 0. Then,

$\begin{align}

& 4-{{k}^{2}}=0 \\

& \Rightarrow {{k}^{2}}=4 \\

& \Rightarrow k=\pm 2 \\

\end{align}$

Thus, when $k=\pm 2$ then the given line will be parallel to the $y-axis$ .

(c) Passing through the origin:

If $\left( k-3 \right)x-\left( 4-{{k}^{2}} \right)y+{{k}^{2}}-7k+6=0$ passes through origin $\left( 0,0 \right)$ . Then, from the first point, we can say that $\left( 0,0 \right)$ will satisfy the equation $\left( k-3 \right)x-\left( 4-{{k}^{2}} \right)y+{{k}^{2}}-7k+6=0$ . Then,

$\begin{align}

& \left( k-3 \right)x-\left( 4-{{k}^{2}} \right)y+{{k}^{2}}-7k+6=0 \\

& \Rightarrow 0-0+{{k}^{2}}-7k+6=0 \\

& \Rightarrow {{k}^{2}}-6k-k+6=0 \\

& \Rightarrow k\left( k-6 \right)-\left( k-6 \right)=0 \\

& \Rightarrow \left( k-6 \right)\left( k-1 \right)=0 \\

& \Rightarrow k=1,6 \\

\end{align}$

Thus, when $k=1,6$ then the given line will pass through the origin.

Note: Although the problem was very easy to solve, the student should apply the concept of coordinate geometry properly to get the right answer. Moreover, the student should avoid making a calculation mistake while solving the question. Here we can also use the concept of slope of parallel line. The slope of the parallel line to the x-axis is 0 and parallel to the y-axis is $\dfrac {{1}}{0}$. For using this concept first we convert the given line in the form of Y = mx + C where m is the slope of the line.

Complete step-by-step solution -

Given:

There is an equation of the straight line $\left( k-3 \right)x-\left( 4-{{k}^{2}} \right)y+{{k}^{2}}-7k+6=0$ . We have to find the value of $k$ for given cases. Before we start solving first look at the following points about a general straight line $L:ax+by+c=0$ :

1. If $L$ passes through a point $\left( h,k \right)$ then, $ah+bk+c=0$ .

2. If $L$ is parallel to the $x-axis$ then, the coefficient of $x$ will be 0 $\left( a=0 \right)$ .

3. If $L$ is parallel to the $y-axis$ then, the coefficient of $y$ will be 0 $\left( b=0 \right)$ .

Now, we will solve each part one by one using the above mentioned points. Then,

(a) Parallel to the $x-axis$ :

If $\left( k-3 \right)x-\left( 4-{{k}^{2}} \right)y+{{k}^{2}}-7k+6=0$ is parallel to the $x-axis$ . Then, from the second point, we can say that the coefficient of $x$ will be 0. Then,

$\begin{align}

& k-3=0 \\

& \Rightarrow k=3 \\

\end{align}$

Thus, when $k=3$ then the given line will be parallel to $x-axis$ .

(b) Parallel to the $y-axis$ :

If $\left( k-3 \right)x-\left( 4-{{k}^{2}} \right)y+{{k}^{2}}-7k+6=0$ is parallel to the $y-axis$ . Then, from the third point, we can say that the coefficient of $y$ will be 0. Then,

$\begin{align}

& 4-{{k}^{2}}=0 \\

& \Rightarrow {{k}^{2}}=4 \\

& \Rightarrow k=\pm 2 \\

\end{align}$

Thus, when $k=\pm 2$ then the given line will be parallel to the $y-axis$ .

(c) Passing through the origin:

If $\left( k-3 \right)x-\left( 4-{{k}^{2}} \right)y+{{k}^{2}}-7k+6=0$ passes through origin $\left( 0,0 \right)$ . Then, from the first point, we can say that $\left( 0,0 \right)$ will satisfy the equation $\left( k-3 \right)x-\left( 4-{{k}^{2}} \right)y+{{k}^{2}}-7k+6=0$ . Then,

$\begin{align}

& \left( k-3 \right)x-\left( 4-{{k}^{2}} \right)y+{{k}^{2}}-7k+6=0 \\

& \Rightarrow 0-0+{{k}^{2}}-7k+6=0 \\

& \Rightarrow {{k}^{2}}-6k-k+6=0 \\

& \Rightarrow k\left( k-6 \right)-\left( k-6 \right)=0 \\

& \Rightarrow \left( k-6 \right)\left( k-1 \right)=0 \\

& \Rightarrow k=1,6 \\

\end{align}$

Thus, when $k=1,6$ then the given line will pass through the origin.

Note: Although the problem was very easy to solve, the student should apply the concept of coordinate geometry properly to get the right answer. Moreover, the student should avoid making a calculation mistake while solving the question. Here we can also use the concept of slope of parallel line. The slope of the parallel line to the x-axis is 0 and parallel to the y-axis is $\dfrac {{1}}{0}$. For using this concept first we convert the given line in the form of Y = mx + C where m is the slope of the line.