Question

find the acute angle between the following pairs of line 2x-y+3=0 and x+3y+7=0

Answer 1

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Answer 2

For a line y = mx + c, m is the Slope of the line.

Given lines

2x - y + 3 = 0

x + 3y + 7 = 0

First line :

⇒2x - y + 3 = 0

⇒2x - y = - 3

⇒2x + 3 = y

⇒ y = 2x + 3

Slope of the line is 2

Second line :

⇒x + 3y = - 7

⇒ 3y = - x - 7

⇒ y = - x/3 - 7/3

Slope of the line is - 1/3

Let x be the angle made by the first line,

Since Slope = 2

tan x = 2

Let y be the angle made by the second line,

Since Slope = - 1/3

tan y = - 1/3

Now angle between the lines is (y-x)

$\tan(A - B) = \frac{ \tan( A) - \tan(B)}{1 - \tan(A) \tan(B) }$

So,

$\tan(y - x) = |\frac{ \tan(y) - \tan(x) }{1 + \tan(x) \tan(y) } |$

$\tan(y - x) = | \frac{ \frac{ - 1}{3} - 2 }{1 - \frac{2}{3} } | \\ \\ \tan(y - x) = | \frac{ \frac{ - 7}{3} }{ \frac{1}{3} } | \\ \\ \tan(y - x) = | - 7| \\ \\ \tan(y - x) = 7$

Therefore, The angle between the two given lines is tan^-1 ( 7)