Question

find the value (s) of m,for which the lines represented by the following pair of linear equations 3x+6y-15=0 and 9x + 18y -m = 0 be coincident​

Answer 1

Answer:

=> m= 45

Step-by-step explanation:

Given that , two lines are coincident.

3x + 6y - 15 = 0   and  9x + 18y - m = 0

To find the value of m, that both lines are coincident.

FORMULA

$\frac{a1} {a2}\neq \frac{b1}{b2} ~~~~~~~~~intersecting\\ \\\frac{a1}{a2}=\frac{b1}{b2}=\frac{c1}{c2}~~coincident\\ \\ \frac{a1}{a2}=\frac{b1}{b2}\neq \frac{c1}{c2}~~parallel$

As per question both lines are coincident,

So, $\frac{a1}{a2}=\frac{b1}{b2}=\frac{c1}{c2}$

=> $\frac{3}{9}=\frac{6}{18}=\frac{-15}{-m} \\ \\\frac{1}{3}=\frac{1}{3}=\frac{15}{m} \\ \\m =15*3\\m= 45$

Therefore m = 45 for the both lines are coincident.

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Answer 2

The value of m is 45, for the lines to be coincident.

Given:

• Pair of two linear equations.
• $3x + 6y - 15 = 0$and $9x + 18y - m = 0$

To find:

• Find the value(s) of m,for which the lines be coincident.

Solution:

Concept/Formula to be used:

Condition of coincident of two lines:

If two linear equation $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$, if

$\bf \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

then the lines are coincident.

Step 1:

Write the coefficients of the equation.

$a_1 = 3 \\ b_1 = 6 \\ c_1= - 15$

and

$a_2 = 9 \\ b_2 = 18\\ c_2= - m$

Step 2:

Put the values in the condition.

$\frac{3}{9} = \frac{6}{18} = \frac{ - 15}{ - m}$

or

$\frac{1}{3} = \frac{1}{3} = \frac{15}{m}$

or

$\bf \red{m = 45 }$

Thus,

m=45, for the lines to be coincident.

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