Question

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Answer 1

$\large\underline{\sf{Given- }}$

Pair of linear equation :-

x + 5y - 7 = 0

and

4x + 20y + k = 0

$\large\underline{\sf{To\:Find - }}$

Value of k when pair of line is coincident.

$\large\underline{\sf{Solution-}}$

We know that

The pair of linear equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 have

$\red{\boxed{ \bf{ \: 1) \: unique \: solution \: when \:   \rm \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}}}}$

$\green{\boxed{ \bf{ \: 2) \: infinite \: solutions \: when \:   \rm \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}}}}$

$\blue{\boxed{ \bf{ \: 3) \: no \: solution \: when \:\rm \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}}}}$

Here,

Given pair of equations,

x + 5y – 7 = 0

4x + 20y + k = 0

Comparing the given two equations with

a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we get  

a₁ = 1

b₁ = 5

c₁ = -7  

a₂ = 4

b₂ = 20

c₂ = k

Now, It is given that lines are coincident. It means system of equations have infinitely many solutions.

So, we know that

System of equations have infinitely many solutions iff

$\rm :\longmapsto\:  \rm \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$

On substituting the values, we get

$\rm :\longmapsto\:\dfrac{1}{4} = \dfrac{5}{20} = \dfrac{ - 7}{k}$

$\rm :\longmapsto\:\dfrac{1}{4} = \dfrac{1}{4} = \dfrac{ - 7}{k}$

So,

$\rm :\longmapsto\: \dfrac{1}{4} = \dfrac{ - 7}{k}$

$\bf\implies \:k = - \: 28$

Hence, Option (b) is correct.