Question

Question 1 Find the values of k for which the line (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.

Class X1 - Maths -Straight Lines Page 233

Answer 1

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) = 0
k = 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) = 0
k = ±2


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

Answer 2

Answer:

Refer to the attachment for your answer