# Thread: Equation of a line...

1. ## Equation of a line...

The given figure represents the lines y = x +1 and y = 3 x -1. Write down the angles which the lines make with the positive direction of x-axis. Hence determine .

2. Originally Posted by snigdha
The given figure represents the lines y = x +1 and y = 3 x -1. Write down the angles which the lines make with the positive direction of x-axis. Hence determine .
Notice where the lines cross the x-axis.

$\displaystyle y=x+1$ crosses at $\displaystyle x=-1$ and $\displaystyle \sqrt{3}x-1$ crosses at $\displaystyle \frac{1}{\sqrt{3}}$.

Since the lines form a triangle with the x-axis, you can find the lenght of the side opposite of $\displaystyle \theta$ using the x-intercepts.

So just find the point where the lines intersect by solving $\displaystyle x+1=\sqrt{3}x-1$. Once you find the point, you should be able to find the lenghts of the sides.

3. The slopes of the lines are the coefficients of the 'x' term.

What are the slopes of the hypotenuse of a 45,45,90 triangle? Of a 60,30,90 triangle?

4. Originally Posted by snigdha
The given figure represents the lines y = x +1 and y = 3 x -1. Write down the angles which the lines make with the positive direction of x-axis. Hence determine .
Hi snigdha,

The line whose equation is $\displaystyle y = x + 1$ has a slope of 1.

The acute angle formed with the positive direction of the x-axis (which is inside the triangle) is $\displaystyle \tan^{-1}(1)={\color{red}45^{\circ}}$

The line whose equation is $\displaystyle y=\sqrt{3}x-1$ has a slope of $\displaystyle \sqrt{3}$.

The acute angle formed with the positive direction of the x-axis is $\displaystyle \tan^{-1}(\sqrt{3})=60^{\circ}$.
Now we're interested in the obtuse angle inside the triangle, so we determine the supplement of $\displaystyle 60^{\circ}$ which is $\displaystyle {\color{red}120^{\circ}}$

Now, you have two angles of the triangle and can determine the value of $\displaystyle {\color{red}\Theta}$