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Miscellaneous Exercise · Q1

Q.Find the values of kk for which the line (k−3)x−(4−k2)y+k2−7k+6=0(k - 3)x - (4 - k^2)y + k^2 - 7k + 6 = 0 is

(a) Parallel to the x-axis,
(b) Parallel to the y-axis,
(c) Passing through the origin.
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The line is parallel to the xx-axis when k=3k=3; parallel to the yy-axis when k=2k=2 or k=−2k=-2; and passes through the origin when k=1k=1 or k=6k=6.

The given line is

(k−3)x−(4−k2)y+(k2−7k+6)=0,(k-3)x - (4-k^2)y + (k^2-7k+6) = 0,

which has the form Ax+By+C=0Ax+By+C=0 with A=k−3A = k-3, B=−(4−k2)B = -(4-k^2), C=k2−7k+6C = k^2-7k+6.

(a) Parallel to the xx-axis

A line parallel to the xx-axis is horizontal, so its slope is 00. For Ax+By+C=0Ax+By+C=0, the slope is −A/B-A/B, which is 00 exactly when the coefficient of xx vanishes (and B≠0B\ne0, so the line doesn't degenerate):

A=k−3=0 ⇒ k=3.A = k-3 = 0 \ \Rightarrow\ k=3.

Check: B=−(4−9)=5≠0B = -(4-9)=5\ne0, so this is valid.

(b) Parallel to the yy-axis

A line parallel to the yy-axis is vertical — it has no yy-term, so the coefficient of yy must vanish (with A≠0A\ne0):

B=−(4−k2)=0 ⇒ k2=4 ⇒ k=2 or k=−2.B = -(4-k^2)=0 \ \Rightarrow\ k^2=4 \ \Rightarrow\ k=2\ \text{or}\ k=-2.

Check the coefficient of xx in each case: for k=2k=2, A=2−3=−1≠0A=2-3=-1\ne0; for k=−2k=-2, A=−2−3=−5≠0A=-2-3=-5\ne0. Both values are valid.

(c) Passing through the origin

A line passes through the origin (0,0)(0,0) exactly when substituting x=0,y=0x=0,y=0 satisfies the equation — i.e. the constant term is zero:

k2−7k+6=0 ⇒ (k−1)(k−6)=0 ⇒ k=1 or k=6.k^2-7k+6=0 \ \Rightarrow\ (k-1)(k-6)=0 \ \Rightarrow\ k=1\ \text{or}\ k=6.

✓Final answer

  1. k=3k=3
  2. k=2k=2 or k=−2k=-2
  3. k=1k=1 or k=6k=6

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