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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 (b) Parallel to the y-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

             (b) Parallel to the y-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 y-axis is x = 0
it is given that these two lines are parallel to each other
Therefore, their slopes are equal
Slope of y=0 is , m' = \infty = \frac{1}{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}=\frac{1}{0}
4-k^2=0
k=\pm2
Therefore, value of k is \pm2

Posted by

Gautam harsolia

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