A line is parallel to the y -axis if the coefficient of y is zero and the coefficient of x is non-zero. Setting 4 - k² = 0 yields k² = 4, so k = 2 or k = -2. For both values, the coefficient of x is k - 3 ≠ 0, so the condition is satisfied.

A line is parallel to the y -axis if the coefficient of y is zero and the coefficient of x is non-zero. Setting 4 - k² = 0 yields k² = 4, so k = 2 or k = -2. For both values, the coefficient of x is k - 3 ≠ 0, so the condition is satisfied.

A line passes through the origin (0, 0) if its constant term is zero, meaning C = 0. Factorising k² - 7k + 6 = 0 gives (k - 1)(k - 6) = 0, which yields k = 1 or k = 6. For both values, the coefficients A and B are not simultaneously zero.

A line passes through the origin (0, 0) if its constant term is zero, meaning C = 0. Factorising k² - 7k + 6 = 0 gives (k - 1)(k - 6) = 0, which yields k = 1 or k = 6. For both values, the coefficients A and B are not simultaneously zero.

Miscellaneous Exercise on Chapter 9 1. Find the values of $k$ for which the line $(k-3)x - (4-k^2)y + k^2 - 7k + 6 = 0$ is (a) Parallel to the $x$ -axis, (b) Parallel to the $y$ -axis, (c) Passing through the origin. 2. Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are $1$ and $-6$ , respectively.

Answer: (a) $k = 3$ , (b) $k = \pm 2$ , (c) $k = 1$ or $k = 6$

Step-by-step solution

Step 1: Analyze the general linear equation

Comparing the given equation with the general equation of a line $Ax + By + C = 0$ , we identify $A = k - 3$ , $B = -(4 - k^2) = k^2 - 4$ , and $C = k^2 - 7k + 6$ .

Step 2: Condition for line parallel to the x-axis

A line $Ax + By + C = 0$ is parallel to the $x$ -axis if the coefficient of $x$ is zero and the coefficient of $y$ is non-zero. Here, $A = k - 3 = 0$ , which gives $k = 3$ . For $k = 3$ , the coefficient of $y$ is $-(4 - 3^2) = 5 \neq 0$ , so the line is parallel to the $x$ -axis.

Step 3: Condition for line parallel to the y-axis

A line is parallel to the $y$ -axis if the coefficient of $y$ is zero and the coefficient of $x$ is non-zero. Setting $4 - k^2 = 0$ yields $k^2 = 4$ , so $k = 2$ or $k = -2$ . For both values, the coefficient of $x$ is $k - 3 \neq 0$ , so the condition is satisfied.

Step 4: Condition for line passing through the origin

A line passes through the origin $(0, 0)$ if its constant term is zero, meaning $C = 0$ . Factorising $k^2 - 7k + 6 = 0$ gives $(k - 1)(k - 6) = 0$ , which yields $k = 1$ or $k = 6$ . For both values, the coefficients $A$ and $B$ are not simultaneously zero.

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