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Find the values of k for which the line (k – 3) x – (4 – k2) y + k2 – 7k + 6 = 0 is (a) Parallel to the x-axis, (b) Parallel to the y-axis,

(a) Here if the line is corresponding to the x-pivot

 

Slant of the line = Slope of the x-pivot

 

It very well may be composed as

 

\[\left( \mathbf{4}\text{ }\text{ }\mathbf{k2} \right)\text{ }\mathbf{y}\text{ }=\text{ }\left( \mathbf{k}\text{ }\text{ }\mathbf{3} \right)\text{ }\mathbf{x}\text{ }+\text{ }\mathbf{k2}\text{ }\text{ }\mathbf{7k}\text{ }+\text{ }\mathbf{6}\text{ }=\text{ }\mathbf{0}\]

We get

NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines image - 47

By additional estimation

\[\mathbf{k}\text{ }\text{ }\mathbf{3}\text{ }=\text{ }\mathbf{0}\]

\[\mathbf{k}\text{ }=\text{ }\mathbf{3}\]\[\mathbf{k}\text{ }=\text{ }\mathbf{3}\]

Thus, assuming the given line is corresponding to the x-hub, the worth of \[\mathbf{k}\text{ }\mathbf{is}\text{ }\mathbf{3}.\]

(b) Here if the line is corresponding to the y-pivot, it is vertical and the slant will be vague

So the slant of the given line

\[\mathbf{k2}\text{ }=\text{ }\mathbf{4}\]

\[\mathbf{k}\text{ }=\text{ }\pm \text{ }\mathbf{2}\]

Thus, assuming the given line is corresponding to the y-hub, the worth of \[\mathbf{k}\text{ }\mathbf{is}\text{ }\pm \text{ }\mathbf{2}.\]