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Find the distance between the parallel lines 3x-y+7=0 and 3x-y+16=0

written 5.6 years ago by teamques10 ★ 69k

modified 3.0 years ago by pedsangini276 • 4.8k

engineering mathematics

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written 5.6 years ago by teamques10 ★ 69k

Solution:

For 3x-y+7=0

$a=3,b=-1,c_{1}=7$

For 3x-y+16=0

$a=3,b=-1,c_{2}=16$

Distance between two parallel lines is

$=\bigg|\frac{c_{2} - c_{1}}{\sqrt{a^{2}+b^{2}}}\bigg| = \bigg|\frac{16-7}{\sqrt{3^{2}+(-1)^{2}}}\bigg|$

$=\bigg|\frac{9}{\sqrt{10}} \bigg|$

$=\frac{9}{\sqrt{10}}. OR 2.846$

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