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  • 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,

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.

Answers (1)

best_answer

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
 

Posted by

Gautam harsolia

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