# Q : 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.

G Gautam harsolia

Given equation of line is
$(k-3)x-(4-k^2)y+k^2-7k+6=0$
and equation of x-axis is $y=0$
it is given that these two lines are parallel to each other
Therefore, their slopes are equal
Slope of $y=0$ is , $m' = 0$
and
Slope of line $(k-3)x-(4-k^2)y+k^2-7k+6=0$  is , $m = \frac{k-3}{4-k^2}$
Now,
$m=m'$
$\frac{k-3}{4-k^2}=0$
$k-3=0$
$k=3$
Therefore, value of k is 3

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