Question

$\huge\rm\underline{\underline{\pink{Question:-}}}$

A square has an area of 25 sq.units is formed by taking two sides as 3x + 4y = k₁ and 3x + 4y = k₂ , then
| k₁ - k₂| is

1] 5

2] 1

3] 25

4] 125


( This question is related to concurrenct of lines in coordinate-geometry . Please answer if u know only. spammers will be reported )




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Answer 1

ANSWER:

• The value of | k₁ - k₂ | = 25

GIVEN:

• Area of square = 25 sq.units

• Two sides are 3x + 4y = k₁ and 3x + 4y = k₂

TO FIND:

• The value of | k₁ - k₂ |

EXPLANATION:

Area of square = s²( where s => side)

s² = 25

s = 5 units

$\boxed{ \large{ \bold {d = \left| \dfrac{k_1 - k_2}{ \sqrt{ a^{2} + b^{2}}} \right| }}}$

a = 3

b = 4

Here we should take the perpendicular distance. As the opposite sides are parallel in square we can take distance as s.

d = s = 5 units

$\sf{5= \left| \dfrac{k_1 - k_2}{ \sqrt{ 3^{2} + 4^{2}}} \right|}$

$\sf{5= \left| \dfrac{k_1 - k_2}{ \sqrt{ 9 + 16}} \right|}$

$\sf{5= \left| \dfrac{k_1 - k_2}{ \sqrt{ 25}} \right| }$

$\sf{5= \dfrac{ |k_1 - k_2| }{5}}$

As modulus is there we can take √25 = 5

$\sf{|k_1 - k_2| = 25}$

Hence the value of | k₁ - k₂ | = 25.

NOTE : REFER ATTACHMENT FOR DIAGRAM.

Answer 2

GIVEN :–

• Area of square = 25 sq. units

• Sides of square is 3x + 4y = k₁ and 3x + 4y = k₂

TO FIND :–

• | k₁ - k₂| = ?

DIAGRAM :–

$\setlength{\unitlength}{20}\begin{picture}(6,6) \put(1,1){\line(1,0){4}}\put(5,1){\line(0,1){4}} \put(1,1){\line(0,1){4}} \put(1,5){\line(1,0){4}} \put(5.5,1.5){\vector(0,1){3.5}} \put(5.5,4.5){\vector(0, - 1){3.5}} \put(3.25,1){\line( - 2,1){0.5}} \put(3.25,1){\line( - 2, - 1){0.5}} \put(3.25,5){\line( - 2,1){0.5}} \put(3.25,5){\line( - 2, - 1){0.5}} \put(1.7,0.25){ \tt 3x + 4y = K_2 }\put(1.7,5.5){ \tt 3x + 4y = K_1 }\put(5.75,3){ \tt Length \:=5\:unit }\put(0.8,0.5){ \bf A }\put(5,0.5){ \bf D }\put(1,5.25){ \bf B }\put(5,5.25){ \bf C }\end{picture}$

SOLUTION :–

• We know that Distance between two parallel lines ax + by + c₁ = 0 & ax + by + c₂ = 0 is –

$\\ \: \bf \: \dag \: \: { \boxed{\bf Distance = \left | \dfrac{ c_{1} - c_{2}}{ \sqrt{ {a}^{2} + {b}^{2} } } \right| }}$

• Now put the values –

$\\ \implies\bf Distance = \left | \dfrac{ k_{1} - k_{2}}{ \sqrt{ {3}^{2} + {4}^{2} } } \right|$

$\\ \implies\bf Distance = \left | \dfrac{ k_{1} - k_{2}}{ \sqrt{ 9 + 16 } } \right|$

$\\ \implies\bf Distance = \left | \dfrac{ k_{1} - k_{2}}{ \sqrt{25} } \right|$

$\\ \implies\bf Distance = \left | \dfrac{ k_{1} - k_{2}}{5} \right|$

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

• Distance between these lines is equal to Length of square.

$\\ \implies\bf Length = \dfrac{ | k_{1} - k_{2} | }{5} \: \: \: \: \: - - - - eq.(1)$

• We also know that –

$\\ \implies \large{ \boxed{\bf Area \: \: of \: \: square = {(Length) }^{2}}}$

• Put the values –

$\\ \implies \bf 25 = {(Length) }^{2}$

$\\ \implies \bf Length = \sqrt{25}$

$\\ \implies \bf Length =5$

• Put in eq.(1) –

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

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

$\\ \implies \large{ \boxed{\bf | k_{1} - k_{2} | = 25 }}$

▪︎ Hence , Option (3) is correct.