Question

Perpendicular distance between 3x + 4y - 5 = 0 and 6x + 8y - 15 = 0 is . .

(A) 1
(B) 1/2.
(C) 22/10
(D) 2​

Answer 1

GIVEN :–

• Line are –

$\\ \: \: \bf {\huge{.}} \: \: 3x + 4y - 5 = 0$

$\\ \implies\bf6x + 8y -15 = 0$

$\\ \: \: \bf {\huge{.}} \: \: 3x + 4y - \dfrac{15}{2} = 0$

TO FIND :–

• Perpendicular distance = ?

SOLUTION :–

• We know that distance between two parallel lines ax+by+c=0 and ax+by+d=0 is –

$\\ \implies{ \boxed{ \red{\bf Distance = \bigg | \dfrac{c - d}{ \sqrt{ {a}^{2} +{b}^{2}}} \bigg| }}}$

• Now , Put the values –

$\\ \implies\bf Distance = \bigg | \dfrac{ - 5-( -15/2) }{ \sqrt{ {3}^{2} +{4}^{2}}} \bigg|$

$\\ \implies\bf Distance = \bigg | \dfrac{ - 5 + 15/2}{ \sqrt{9+16}} \bigg|$

$\\ \implies\bf Distance = \bigg | \dfrac{(15 - 10)/2}{ \sqrt{25}} \bigg|$

$\\ \implies\bf Distance = \bigg | \dfrac{(15 - 10)/2}{5}\bigg|$

$\\ \implies\bf Distance = \bigg | \dfrac{5/2}{5}\bigg|$

$\\ \implies\bf Distance = \bigg | \dfrac{5}{5 \times 2}\bigg|$

$\\ \implies\bf Distance = \bigg | \cancel\dfrac{5}{10}\bigg|$

$\\ \implies\bf Distance = \bigg | \dfrac{1}{2}\bigg|$

$\\ \implies \red{ \boxed{\bf Distance = \dfrac{1}{2}}}$

• Hence , Option (B) is correct.

Answer 2

Option (B) 1/2

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Given :-

• Lines

1. 3x + 4y - 5 = 0
2. 6x + 8y - 15 = 0

To find :-

• Perpendicular Distance between them.

Solution :-

Making the line 2 same as line 1.

${\sf{ \ratio\longmapsto{6x + 8y - 15 = 0}}}$

${\sf{ \ratio\longmapsto{ \cfrac{ 6x + 8y - 15}{2} = 0}}}$

${\sf{ \ratio\longmapsto{ \cfrac{ 6x }{2}+ \cfrac{ 8y}{2} - \cfrac{ 15}{2}= 0}}}$

${ \boxed{\sf{ \ratio\longmapsto{ 3x + 4y - \cfrac{ -15}{2}= 0}}}} ..........(eq \:2)$

So, As we have seen that slopes of the lines are same so Perpendicular distance between can be found by given below formula :-

${\sf{ \underline{\boxed{\sf{ Perpendicular \: Distance = \dfrac{|c_1 - c_2 |}{ \sqrt{ {a}^{2} + {b}^{2} } }}}}}}$

Comparing equation with :-

(1) ax + by + c_1 with 3x + 4y - 5 = 0 we get,

• a = 3
• b = 4
• c1 = -5

(2) ax + by + c_2 with 3x + 4y - 15/2 = 0 we get,

• a = 3
• b = 4
• c = -15/2

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So, substituting the values in formula :-

${ \ratio\implies{\sf{ Perpendicular \: Distance = \dfrac{|c_1 - c_2 |}{ \sqrt{ {a}^{2} + {b}^{2} } }}}}$

${ \ratio\implies{\sf{ Perpendicular \dfrac{ \bigg| - 5 - \bigg( - \cfrac{ 15}{2} \bigg) \bigg|}{ \sqrt{ {3}^{2} + {4}^{2} } }}}}$

${ \ratio\implies{\sf{ \dfrac{ \bigg| - 5 + \cfrac{ 15}{2} \bigg|}{ \sqrt{ {3}^{2} + {4}^{2} } }}}}$

${ \ratio\implies{\sf{ \dfrac{ \bigg| \cfrac{ - 5 \times 2}{1 \times 2} + \cfrac{ 15}{2} \bigg|}{ \sqrt{ {3}^{2} + {4}^{2} } }}}}$

${ \ratio\implies{\sf{ \dfrac{ \bigg| \cfrac{ - 10}{2} + \cfrac{ 15}{2} \bigg|}{ \sqrt{ {3}^{2} + {4}^{2} } }}}}$

${ \ratio\implies{\sf{ \dfrac{ \bigg| \cfrac{ - 10 + 15}{2} \bigg|}{{ \sqrt{9 + 16}} }}}}$

${ \ratio\implies{\sf{ \dfrac{ \bigg| \: \: \cfrac{ 5}{2} \: \: \bigg|}{{ \sqrt{25} } }}}}$

${ \ratio\implies{\sf{ \dfrac{ \cfrac{ 5}{2}}{ \: \: \: \: \: \: 5\: \: \: \: \: \: \: } }}}$

${ \ratio\implies{\sf{ \cfrac{ 5}{2} \: \div \cfrac{5}{1} }}}$

${ \ratio\implies{\sf{ \cfrac{ 5}{2} \: \times \cfrac{1}{5} }}}$

${ \ratio\implies{\sf{ \cfrac{ 5}{2 \times 5} \: }}}$

${ \ratio\implies{\sf{ \cfrac{ 5}{10} \: }}}$

${ \underline{ \boxed{ \green{ \ratio\implies{\sf{ Perpendicular \: Distance = \cfrac{ 1}{2} \: }}}}}}$