Question

Find k if the graph is parallel for the following: 2x-ky=9, 6x-9y=18

Answer 1

Answer:

For the lines ax+by=c and Ax+By=C to be parallel. We must have

a/A=b/B≠c/C

Therefore

2/6=k/9

k=3

Answer 2

Answer:

k = 3

Step-by-step explanation:

Two lines are given ,

2x - ky = 9  and   6x - 9y = 18

We can rewrite it as given below

2x + (-k)y + (-9) = 0   and  6x + (-9)y + (-18) = 0

FORMULA

a1 x + b1y + c1 = 0

a2x + b2y + c2 = 0

°    $\frac{a1}{a2} \neq \frac{b1}{b2}$              ( intersecting)

°   $\frac{a1}{a2} =\frac{b1}{b2} =\frac{c1}{c2}$       ( coincident )

°    $\frac{a1}{a2} =\frac{b1}{b2} \neq \frac{c1}{c2}$      ( parallel )

As per question, the two given lines are parallel.

So, $\frac{a1}{a2} =\frac{b1}{b2}\neq \frac{c1}{c2}$ satisfy it.

∴ $\frac{2}{6}=\frac{-k}{-9}\neq \frac{-9}{-18}$

  $\frac{k}{9}=\frac{2}{6} \\\\\frac{k}{9}=\frac{1}{3}\\\\ \frac{k}{1} =\frac{9}{3}\\ \\ k=3$

Therefore , the value of K = 3 , then the both lines are parallel.

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