Q1.find the acute angle between the following pairs of line 2x-y+3=0 and x+3y+7=0.
Answers
Step-by-step explanation:
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)
So,
Therefore, The angle between the two given lines is tan^-1 ( 7).
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}\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} \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).