Find the value of k for which the line (k-3)x | 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 $x$-axis,its slope must be $0$. Rewriting t

Vedclass · Accepted Answer

$k = 3$

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 $x$-axis,its slope must be $0$. Rewriting the equation in the form $y = mx + c$: $(4-k^2)y = (k-3)x + (k^2 - 7k + 6)$ $y = \frac{k-3}{4-k^2}x + \frac{k^2 - 7k + 6}{4-k^2}$. For the line to be parallel to the $x$-axis,the slope $m = \frac{k-3}{4-k^2}$ must be $0$. This implies $k-3 = 0$,so $k = 3$. However,we must ensure the coefficient of $y$ is not zero. If $k=3$,the coefficient of $y$ is $-(4-3^2) = -(4-9) = 5 
eq 0$. Thus,the value of $k$ is $3$.

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 $x$-axis.

Similar Questions

The points $(1, 3)$ and $(5, 1)$ are the opposite vertices of a rectangle. The other two vertices lie on the line $y = 2x + c$. Then the value of $c$ is:

The vertex $A$ of a triangle lies on the lines $x+y=1$ and $2x+3y=6$. If the orthocentre of the triangle is $O\left(\frac{3}{7}, \frac{22}{7}\right)$,then the equation of $OA$ in the normal form is

Find the equation of the line that bisects the line segment joining the points $(5, 3)$ and $(4, 4)$ and makes an angle of $45^{\circ}$ with the positive direction of the $x$-axis.

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

The line passing through the points $(a, b)$ and $(-a, -b)$ passes through which of the following points?

Vedclass Test Series

Exam Paper Generator

Online Exam Module