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# Find the value of k for which the line (k - 3) x - (4 - k ^{2}) y + k ^{2} - 7k + 6 = 0 is

(a) Parallel to x-axis (b) Parallel to y-axis

(c) Passing through the origin

**Solution:**

The given equation of the line is

(k - 3)x - (4 - k^{2})y + k^{2} - 7k + 6 = 0 ....(1)

**(a)** Then given line can be written as

(k - 3)x + k^{2} - 7k + 6 = (4 - k^{2}) y

y = (k - 3)/(4 - k^{2})x + (k^{2} - 7k + 6)/(4 - k^{2})

Which is of the form y = mx + c

Slope of the given line = (k - 3)/(4 - k^{2})

Slope of the x-axis = 0

If the given line is parallel to the x-axis then,

Slope of the given line = Slope of the x-axis.

⇒ (k - 3)/(4 - k^{2}) = 0

⇒ k - 3 = 0

⇒ k = 3

Thus, the given line is parallel to x-axis, then the value of k = 3.

**(b)** If the given line is parallel to the y-axis, it is vertical.

Hence, its slope will be undefined.

The slope of the given line is = (k - 3)/(4 - k^{2})

Now, (k - 3)/(4 - k^{2}) is defined at k^{2} = 4

⇒ k^{2} = 4

⇒ k = ± 2

Thus, if the given line is parallel to the y-axis, then the value of k = ± 2

**(c)** If the given line is passing through the origin, then point (0, 0) satisfies the given equation of the line.

(k - 3)(0) - (4 - k^{2})(0) + k^{2} - 7k + 6 = 0

⇒ k^{2} - 7k + 6 = 0

⇒ k^{2} - 6k - k + 6 = 0

⇒ (k - 6)(k - 1) = 0

⇒ k = 6 or k = 1

Thus, if the given line is passing through the origin, then the value of k is either 1 or 6

NCERT Solutions Class 11 Maths Chapter 10 Exercise ME Question 1

## Find the value of k for which the line (k - 3) x - (4 - k ^{2}) y + k ^{2} - 7k + 6 = 0 is (a) Parallel to x-axis (b) Parallel to y-axis (c) Passing through the origin.

**Summary:**

a) The given line is parallel to x-axis, then the value of k = 3

b) The given line is parallel to the y-axis, then the value of k = ± 2

c) The given line is passing through the origin, then the value of k is either 1 or 6

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