Miscellaneous Exercise on Chapter 10Find the values of k for which the

1.	Miscellaneous Exercise on Chapter 10Find the values of k for which the line (k-3) x (4-)y -7k+6isa) Parallel to the x-axis,Parallel to the y-axis,Passing through the origin.
Answer» Equation of line is (k -3)x -(4 - k²)y+ (k²-7k+6) = 0 (a) we know, when any line parallel to x-axis then slope of line is equal to zero. slope of given line = (k-3)/(4-k²) 0 = (k-3)/(4 - k²)(k -3) = 0k = 3 (b) when line parallel to y - axis then, slope of line = ∞ = 1/0 here, Slope of line = (k-3)/(4-k²)1/0 = (k-3)/(4-k²)(4-k²) = 0k = ±2 (c) when line passing through origin Then, x = 0, and y = 0 So, (k-3)0 -(4-k²)0 + k²-7k+6 = 0k²-7k+6 = 0k²-6k-k+6 = 0k(k-6)-(k-6)=0(k-1)(k-6)=0k=1,6

Miscellaneous Exercise on Chapter 10Find the values of k for which the line (k-3) x (4-)y -7k+6isa) Parallel to the x-axis,Parallel to the y-axis,Passing through the origin.

Answer»

Equation of line is (k -3)x -(4 - k²)y+ (k²-7k+6) = 0

(a) we know, when any line parallel to x-axis then slope of line is equal to zero.

slope of given line = (k-3)/(4-k²) 0 = (k-3)/(4 - k²)(k -3) = 0k = 3

(b) when line parallel to y - axis then, slope of line = ∞ = 1/0 here, Slope of line = (k-3)/(4-k²)1/0 = (k-3)/(4-k²)(4-k²) = 0k = ±2

(c) when line passing through origin Then, x = 0, and y = 0 So, (k-3)0 -(4-k²)0 + k²-7k+6 = 0k²-7k+6 = 0k²-6k-k+6 = 0k(k-6)-(k-6)=0(k-1)(k-6)=0k=1,6