Find the value of k for which the line (k-3) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The given equation of the line is $(k-3) x - (4-k^2) y + k^2 - 7k + 6 = 0$. If a line is parallel to the $y$-axis,it is a vertical line of the for

Vedclass · Accepted Answer

$\pm 2$

Vedclass · Answer

(C) The given equation of the line is $(k-3) x - (4-k^2) y + k^2 - 7k + 6 = 0$. If a line is parallel to the $y$-axis,it is a vertical line of the form $x = c$. For the equation to represent a vertical line,the coefficient of $y$ must be zero,and the coefficient of $x$ must be non-zero. Setting the coefficient of $y$ to zero: $-(4-k^2) = 0$ $\Rightarrow 4-k^2 = 0$ $\Rightarrow k^2 = 4$ $\Rightarrow k = \pm 2$. Now,check the coefficient of $x$ for these values: If $k = 2$,the coefficient of $x$ is $(2-3) = -1 
eq 0$. If $k = -2$,the coefficient of $x$ is $(-2-3) = -5 
eq 0$. Since both values result in a valid vertical line,the value of $k$ is $\pm 2$.

Find the value of $k$ for which the line $(k-3) x - (4-k^2) y + k^2 - 7k + 6 = 0$ is parallel to the $y$-axis.

Similar Questions

The $y$-intercept of the line passing through $A(6, 1)$ and perpendicular to the line $x - 2y = 4$ is:

If each of the points $(a, 4)$ and $(-2, b)$ lies on the line joining the points $(2, -1)$ and $(5, -3)$,then the point $(a, b)$ lies on the line:

If the three points $(3q, 0)$, $(0, 3p)$, and $(1, 1)$ are collinear, then which one is true?

The equations of the diagonals of the square formed by the lines $x = 0,$ $y = 0,$ $x = 1,$ and $y = 1$ are

The equation of the line,which bisects the line segment joining the points $(2, -19)$ and $(6, 1)$ and is perpendicular to the line joining the points $(-1, 3)$ and $(5, -1)$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module