Q:find thevalues of k for which the line (k-3)x-(4-k^2)y+k^2-7k+6=0 A):parallel to x axis B):parallel to y axis Could you please help me with this question?

For two lines to be parallel slopes should be equal .

 

Slope of the given line= -(coefficient of X)/(coefficient of Y)=-(k-3)/-(4-k^2)=(k-3)/(4-k^2P)

 

A) Slope of X axis =0

Therefor for the line to be parallel to X axis Equte the slope to 0.

(k-3)/(4-k^2)=0

therefor K=3

 

B) Slope of Y axis=?

Therefor for the line to be parallel to Y axis Equte the slope to ?.

(k-3)/(4-k^2)=?

4-k^2=0

k=±2