Question

find the angle between the lines 2x+y+3=0 and x+3y+4=0​

Answer 1

Answer:

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) }tan(A−B)=

1−tan(A)tan(B)

tan(A)−tan(B)

So,

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

1+tan(x)tan(y)

tan(y)−tan(x)

∣

\begin{gathered} \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\end{gathered}

tan(y−x)=∣

1−

3

2

3

−1

−2

∣

tan(y−x)=∣

3

1

3

−7

∣

tan(y−x)=∣−7∣

tan(y−x)=7

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

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