Question

Let x-3=y and ky-x+6=0 represent a pair of parallel lines then k =​

Answer 1

Step-by-step explanation:

Given that:

x-3=y and ky-x+6=0

To find:

Value of k for which lines are parallel

Solution:

Two lines

$a_1x + b_1y + c_1 = 0 \\ \\ a_2x + b_2y + c_2 = 0$

are parallel, if

$\frac{a_1}{a_2} = \frac{b_1}{b_2} \not= \frac{c_1}{c_2}$

Here ,first write these equations in standard form

$x - y - 3 = 0 \\ \\ - x + ky + 6 = 0 \\ \\ here \\ \\ a_1 = 1 \: \: b_1 = - 1 \: \: c_1 = - 3 \\ \\ a_2 = - 1 \: \: b_2 = k \: \: c_2 = 6$

Write these values in the condition

$\frac{1}{ - 1} = \frac{ - 1}{k} \not= \frac{ - 3}{6 } \\ \\ \frac{1}{ - 1} = \frac{ - 1}{k} \not= \frac{ - 1}{2}$

from first two

$k = 1 \\ \\ from \: last \: two \\ \\ k \not= 2$

Hope it helps you.

Answer 2

Answer:

k=1

is your answer.......