Question

Distance between the lines 5x + 3y – 7 = 0 and
15x + 9y + 14 = 0 is​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

The distance between the lines is $d=\frac{7}{3\sqrt{34}}$  unit.

Step-by-step explanation:

Given : Lines $5x + 3y -7 = 0$ and $15x + 9y + 14 = 0$

To find : Distance between the lines ?

Solution :

The formula of distance between two lines $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ is given by,

$d=|\frac{c_1}{\sqrt{a_1^2+b_1^2}}|+|\frac{c_2}{\sqrt{a_2^2+b_2^2}}|$

On comparing, $a_1=5,b_1=3,c_1=-7,a_2=15,b_2=9,c_2=14$

Substitute the values,

$d=|\frac{-7}{\sqrt{5^2+3^2}}|+|\frac{14}{\sqrt{15^2+9^2}}|$

$d=|\frac{-7}{\sqrt{25+9}}|+|\frac{14}{\sqrt{225+81}}|$

$d=|\frac{-7}{\sqrt{34}}|+|\frac{14}{\sqrt{306}}|$

$d=|\frac{-7}{\sqrt{34}}|+|\frac{14}{3\sqrt{34}}|$

$d=|\frac{-21+14}{3\sqrt{34}}|$

$d=|\frac{-7}{3\sqrt{34}}|$

$d=\frac{7}{3\sqrt{34}}$

Therefore, the distance between the lines is $d=\frac{7}{3\sqrt{34}}$  unit.

#Learn more

How to find the perpendicular distance between the parallel lines 4x-3y+7=0 and 12x-3y+5=0​

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