Question

find the distance between the parallel lines 3x - 4y + 7 = 0 and 3x - 4y + 6 = 0​

Answer 1

Answer:

1 unit is ur answer . hope it helps

Answer 2

☆Given Question :-

• Find the distance between the parallel lines 3x - 4y + 7 = 0 and 3x - 4y + 6 = 0.

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☆Required Answer:-

$\sf \: \dfrac{1}{5} \: units$

☆Explanation:-

$\large\sf{Given:-}</p><p>$

• Two parallel lines 3x - 4y + 7 = 0 and 3x - 4y + 6 = 0

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

• Distance between the parallel lines 3x - 4y + 7 = 0 and 3x - 4y + 6 = 0

☆ Formula used :-

Let us consider two parallel lines ax + by + c = 0 and ax + by + d = 0, then distance between parallel lines is given by

$\bf \:Distance = \dfrac{ |c - d| }{ \sqrt{ {a}^{2} + {b}^{2} } }$

$\large\sf{Processing:- }$

Two lines are

$\sf \: 3x - 4y + 7 = 0 ⟼ \: (1)$

$\sf \: 3x - 4y + 6 \: = 0 \: ⟼ \: (2)$

So, Distance between two parallel lines is given by

$\bf \:Distance = \dfrac{ |c - d| }{ \sqrt{ {a}^{2} + {b}^{2} } }$

On substituting the values of a = 3, b = - 4, c = 7, d = 6, we get

$\sf \:Distance = \dfrac{ |7 - 6| }{ \sqrt{ {3}^{2} + {( - 4)}^{2} } }$

$\sf \:Distance = \dfrac{ |1| }{ \sqrt{ 9 + 16 } }$

$\sf \:Distance = \dfrac{ 1}{ \sqrt{25 } } =\dfrac{1}{5} \: units$

$\large{\boxed{\boxed{\sf{Hence, distance \: between \: them \: = \dfrac{1}{5} \: units }}}}$

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