Checking nature of angles of a triangle given the equations of the three lines that form a triangle Suppose we have three lines $\ell_i=a_ix+b_iy=c_i$, $i=1,2,3$ and we are given that they form a triangle. I need to find
which angles are acute and which are obtuse without plotting the lines
graphically (not even roughly). What would be the best method to do so?
I vaguely remember a method I saw in a book but can't find it now. Most probably it told to find out the $\tan$ values of the angles to check if they are acute or obtuse. I don't think that's correct however, because there can be two angles between a pair of lines. One acute and one obtuse. So I'm a bit confused.
N.B. I am in high school and am not familiar with inner products yet.
 A: The nature of the triangle can be determined without explicitly finding its vertices.  
Let $s_{ij} = (a_ia_j+b_ib_j)(a_ib_j-b_ia_j)$ and examine the signs of $s_{12}$, $s_{23}$ and $s_{31}$. (The order of the indices of the last one is important.)


*

*If they are all positive or all negative, then the triangle is acute.

*If one of them is zero and the others are either both positive or both negative, it is a right trangle.

*If one is negative and the other two positive, or vice-versa, the triangle is obtuse. If the odd man out is $s_{ij}$, the obtuse angle is at the intersection of lines $\ell_i$ and $\ell_j$.  


Any other combination of signs means that the three lines do not form a triangle.  
Notice that the $c_i$’s don’t enter into this solution. This makes sense: If you vary $c_i$ while holding $a_i$ and $b_i$ fixed, you get a family of parallel lines. The resulting triangles are all similar, so their nature depends only on the directions of the three lines.

Why does this work? (Rather long)  
There’s a lot of detail below, but the key to it all is that doubling the angles lets us “factor out” the direction in which we move along the lines.
Suppose we have vectors $\mathbf u=\langle u_1,u_2\rangle$ and $\mathbf v=\langle v_1,v_2\rangle$. Their dot product is defined as $$\mathbf u\cdot\mathbf v = \|\mathbf u\|\|\mathbf v\|\cos\theta = u_1v_1+u_2v_2.$$ Note that the dot product is symmetric: $\mathbf u\cdot\mathbf v = \mathbf v\cdot\mathbf u$. Also, if neither vector is zero, the sign of their dot product is determined by the angle $\theta$ between them: positive if $0\le\theta\lt\frac\pi2$, zero if $\theta=\frac\pi2$ and negative if $\frac\pi2\lt\theta\le\pi$. (This property of the dot product is used in Aretino’s answer.)  
The (signed) area of the parallelogram with sides defined by $\mathbf u$ and $\mathbf v$ is $$\|\mathbf u\|\|\mathbf v\|\sin\theta = \det\left[\matrix{u_1&u_2\\v_1&v_2}\right] = u_1v_2-u_2v_1.$$ If you’re familiar with the cross product, you might recognize this expression as the $z$-component of $\mathbf u\times\mathbf v$, treating them as three-dimensional vectors. Here, too, assuming that neither vector is zero, the sign is determined by the angle between them: positive if a counterclockwise rotation takes $\mathbf u$ onto $\mathbf v$; zero if they’re parallel or antiparallel; negative if the rotation is clockwise. That is, the sign of this value encodes the relative orientation of the two vectors. An immediate consequence is that this expression is antisymmetric in $\mathbf u$ and $\mathbf v$—swapping them flips the sign of the result. In the following, we’re not really interested in the magnitude of this value, only its sign, as that gives us a way to determine the relative orientation of two vectors.  
If we multiply these two together, we get $$(u_1v_1+u_2v_2)(u_1v_2-u_2v_1) = (\|\mathbf u\|\|\mathbf v\|\cos\theta)(\|\mathbf u\|\|\mathbf v\|\sin\theta) = \frac12\|\mathbf u\|^2\|\mathbf v\|^2\sin{2\theta}.$$ Call this $S(\mathbf u,\mathbf v)$ to reduce clutter. Note that this, too, is antisymmetric, i.e., $S(\mathbf u,\mathbf v)=-S(\mathbf v,\mathbf u)$. It also has the easily-verified property that $S(\mathbf u,\mathbf v)=S(-\mathbf u,\mathbf v)=S(\mathbf u,-\mathbf v)=S(-\mathbf u,-\mathbf v)$, that is, that you can reverse either or both of the vectors without changing the value of $S$. It’s also rotationally invariant—rotating both vectors by the same angle leaves $S$ unchanged.  
We start as in Aretino’s answer: Let $P_{ij}$ be the intersection of $\ell_i$ and $\ell_j$ with corresponding angle $\alpha_{ij}$ of the triangle. Consider the vectors $\mathbf u=P_{23}-P_{12}$ and $\mathbf v=P_{31}-P_{12}$, which represent the two sides of the triangle that meet at $P_{12}$. That is, $\mathbf u$ is a directed segment of $\ell_2$ and $\mathbf v$ a directed segment of $\ell_1$. The dot product $\mathbf u\cdot\mathbf v$ will be positive if $\alpha_{12}$ is acute, zero if it’s a right angle and negative if obtuse. We can make a similar argument for each of the other two vertices.  
Assume now that the triangle is oriented such that traversing its vertices in the order $P_{12}$, $P_{23}$, $P_{31}$ takes you in a counterclockwise direction, and at each vertex choose $\mathbf u$ and $\mathbf v$ so that we go counterclockwise from $\mathbf u$ to $\mathbf v$. With these conditions, $S(\mathbf u,\mathbf v)$ will also be positive, zero or negative, respectively, as the corresponding angle of the triangle is acute, right or obtuse. Reversing the direction in which the vectors point doesn’t change $S$, so we can cut down on the proliferation of vectors by only using ones that follow the order of the vertices: $P_{23}-P_{12}$, $P_{31}-P_{23}$ and $P_{12}-P_{32}$. If the vertices are instead arranged in a clockwise direction, this amounts to swapping $\mathbf u$ and $\mathbf v$ at each vertex, which flips the sign of $S(\mathbf u, \mathbf v)$ at all three. Putting all of this together we get the three rules at the top, albeit in terms of vectors that represent the known sides of the triangle.  
Since we care only about the sign of $S(\mathbf u,\mathbf v)$ we don’t really need the lengths of the sides of the triangles, we just need vectors that point in the same directions. We can read those off from the equations of the lines: both $\langle-b_i,a_i\rangle$ and $\langle b_i,-a_i\rangle$ are parallel to $\ell_i$. It’s not immediately obvious which of these is the right one to use, but because we can negate either or both of the arguments to $S$ without changing its value, it doesn’t matter. Setting $s_{ij}=S(\langle-b_i,a_i\rangle,\langle-b_j,a_j\rangle)$ and expanding gives us the result at the top.   
We can also use vectors perpendicular (normal) to the sides—$\langle a_i,b_i\rangle$ or their negations—instead of vectors parallel to the sides since $\langle a_i,b_i\rangle$ is $\langle b_i,-a_i\rangle$ rotated 90 degrees and we can negate the arguments to $S$ at will. I like to have a geometric view to back up the algebra, so assume that we got lucky: the triangle is oriented counterclockwise and the equations of the lines are such that all of the $\mathbf{n}_i$ point outwards, as in the diagram below.

In this configuration, the angle $\theta_{ij}$ between vectors $\mathbf{n}_i$ and $\mathbf{n}_j$ can easily be seen to be the complementary angle to $\alpha_{ij}$, namely $\pi-\alpha_{ij}$. (See the perpendicular bisector theorem for triangles if you want something more rigorous.) This flips the sign of $\sin{2\theta}$, but we’ve also reversed their relative orientation, so the sign of $S$ for $\alpha_{ij}$ is unchanged by using normals. Thus, the rules for determining the nature of a counterclockwise triangle using outward-pointing normals are the same as before.   
Now, what happens if we reverse one of these vectors, say $\mathbf{n}_1$? This adds $\pi$ to the angles between it and its neighbors, which changes the sign of both the sine and cosine of these angles, so the sign of their product is unchanged. Reversing two of the vectors is the same as reversing one and rotating by $\pi$, while reversing all of them is the same as rotating them all by $\pi$. Rotating all of the vectors by the same amount doesn’t change the angles between them, so the signs of the $s_{ij}$ are unchanged by any of these reversals. If the triangle is oriented clockwise, we’ve already seen that this just changes the sign of all of the $s_{ij}$. Thus, the rules for determining the type of triangle also hold if we work with vectors normal to the lines, and it doesn’t matter which of the two normal directions we choose for each.  
As a final note, you can dig a bit deeper to determine if the triangle is isosceles or equilateral by using the fact that $\tan\theta_{ij}={\sin{\theta_{ij}}\over\cos{\theta_{ij}}}={a_ib_j-b_ia_j\over a_ia_j+b_ib_j}$.  
A: So we have three lines $\ell_i=a_ix+b_iy=c_i$, $i=1,2,3$ and we are given that they form a triangle. First we note that the angle beween two lines is completely determined by the $a_i$'s and $b_i$'s. 
We start by finding the vertices of our triangle, i.e. the three intersection points of the lines: $A_{12}$, $A_{23}$, $A_{13}$, where $A_{ij}$ is the point where $\ell_i$ and $\ell_j$ intersect. Now in the picture below you see how we can find $\alpha$ using $a_1$ and $b_1$. $\beta$ is found in a similar way. This method can also be used for the other two angles. 

A: You may:


*

*Find the coordinates of the vertices $A,B,C$;

*Compute $AB^2,AC^2,BC^2$ through the Pythagorean theorem;

*Check if $AB^2+AC^2$ is greater or less than BC^2. That gives you enough information for $\widehat{BAC}$ to be an acute, right or obtuse angle.

A: Suppose we have a triangle $\triangle ABC$.
If we construct a line through vertex $A$ parallel to side $BC$, we will have three lines through a single point forming six angles between "adjacent" rays. 
Three of these angles are equal to the other three, and are also equal to
the interior angles of $\triangle ABC$.
If we apply the inverse tangent formula to the slopes of the three given lines, 
We will get three directional angles in the range from $-\frac\pi2$ ($-90$ degrees) to $\frac\pi2$ ($90$ degrees).
If we consider these directional angles in sequence from least to greatest, and take the differences between the consecutive pairs,
we get two of the interior angles of
$\triangle ABC$.
And of course the third angle is $180$ degrees minus the sum of the other two. 
A: 1) Find the intersection point of every pair of lines: this will give you the three vertices $A$, $B$, $C$ of the triangle.
2) For vertex $A$ compute $(B-A)\cdot(C-A)$, that is 
$$(B_x-A_x)(C_x-A_x)+(B_y-A_y)(C_y-A_y).$$ 
If the result is positive, angle $A$ is acute; if the result is negative, angle $A$ is obtuse; if it vanishes, the angle is right.
3) Repeat for vertices $B$ and $C$.
A: $(a_i, b_i)$ is a vector normal to the the line.
$(a_i, b_i)\cdot(a_j,b_j) = a_ia_j + b_ib_j$ is the "dot" or inner product between two vectors.  And the sign of the dot product is related to the cosine of the angle of intersection.  If it is positive the angle is acute, if it is negative, the angle is obtuse, and if is 0 the angle is right.
