Question

$\huge{\mathcal{\underline{\green{Question:-}}}}$

If the lines x + y + k₁ = 0 and k₂x -5y - 5 = 0 are identical then the value of | k₂ + k₁ | is

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Answer 1

CONCEPT :

Two straight lines are said to be identical straight lines When the coefficients of two straight lines are Proportional.

EXPLANATION :

Suppose

lx + ny + f = 0 &

Lx + Ny + F = 0

be two straight lines.

Then above two straight lines are said to be identical straight lines When l/L = n/N = f/F

TO FIND :

If the lines x + y + k₁ = 0 and k₂x -5y - 5 = 0 are identical then the value of | k₂ + k₁ |

CALCULATION :

Since the lines x + y + k₁ = 0 and k₂x -5y - 5 = 0 are identical

So coefficients of two straight lines are Proportional.

∴ 1/k₂ = 1/-5 = k₁ / - 5

Now 1/k₂ = 1/-5 gives k₂ = - 5

Also 1/-5 = k₁ / - 5 gives k₁ = 1

Hence | k₂ + k₁ | = | - 5 + 1 | = | - 4 | = 4

Answer 2

GIVEN :–

• lines x + y + k₁ = 0 and k₂x -5y - 5 = 0 are identical.

TO FIND :–

• | k₂ + k₁ | = ?

SOLUTION :–

• We know that two lines a₁x + b₁y + c₁ = 0 & a₂x + b₂y + c₂ = 0 are identical then –

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$\bf \implies \large { \boxed{ \bf \dfrac{ a_{1}}{a_{2}} = \dfrac{ b_{1}}{b_{2}} = \dfrac{ c_{1}}{c_{2}}}}$

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• Here –

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$\bf \: \blacktriangleright \: \: a_{1}=1 , b_{1}=1,c_{1} = k_{1}$

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$\bf \: \blacktriangleright \: \: a_{2}=k_{2} , b_{2}=-5 ,c_{2} = -5$

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• So that –

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$\bf \implies \dfrac{1}{k_{2}} = \dfrac{1}{ - 5} = \dfrac{ k_{1}}{ - 5}$

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• Now Let's use $\bf \:\: { \underbrace{ \dfrac{1}{k_{2}} = \dfrac{1}{ - 5}}}= \dfrac{ k_{1}}{ - 5} \: \:$ –

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$\bf \implies \dfrac{1}{k_{2}} = \dfrac{1}{ - 5} \: \:$

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$\bf \implies \large{ \boxed{ \bf k_{2}= - 5}}\: \:$

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• Now Let's use $\bf \:\: { \dfrac{1}{k_{2}} = { \underbrace{ \dfrac{1}{ - 5}= \dfrac{ k_{1}}{ - 5}}}} \: \:$ –

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$\bf \implies \dfrac{1}{ - 5} = \dfrac{k_{1}}{ - 5} \: \:$

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$\bf \implies \large{ \boxed{ \bf k_{1}=1}}\: \:$

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• Hence –

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$\bf \implies | k_{1} + k_{2} | = | - 5 + 1|$

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$\bf \implies | k_{1} + k_{2} | = | - 4|$

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$\bf \implies \large { \boxed{ \bf | k_{1} + k_{2} | =4}}$

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$\rule{220}{4}$