algebra-precalculusanalytic geometrycoordinate systemsdeterminantgeometry
I tried using below stated amazing formula:
Then the area of the triangle that these lines will enclose is given by the magnitude of :
$$\frac{det\begin{bmatrix}a_1 & b_1 & c_1\\a_2 & b_2 & c_2\\a_3 & b_3 & c_3\end{bmatrix}^2}{2C_1C_2C_3}$$
[Where $C_1,C_2,C_3$ are the co-factors of $c_1,c_2,c_3$ respectively in the above matrix.] ref.
But couldn't get the desired answer. System of linear questions are as follows.
$$2x+3y+-6=0$$
$$3x+2y+6=0$$
$$3x+-3y+6=0$$
I end up getting decimal answers!
Clearly, we can scale the coefficients of a given linear equation by any (non-zero) constant and the result is unchanged. Therefore, by dividing-through by $\sqrt{a_i^2+b_i^2}$, we may assume our equations are in "normal form":
$$\begin{align} x \cos\theta + y \sin\theta - p &= 0 \\ x \cos\phi + y \sin\phi - q &= 0 \\ x \cos\psi + y \sin\psi - r &= 0 \end{align}$$
with $\theta$, $\phi$, $\psi$ and $p$, $q$, $r$ (and $A$, $B$, $C$ and $a$, $b$, $c$) as in the figure:
Then
$$C_1 = \left|\begin{array}{cc} \cos\phi & \sin\phi \\ \cos\psi & \sin\psi \end{array} \right| = \sin\psi\cos\phi - \cos\psi\sin\phi = \sin(\psi-\phi) = \sin \angle ROQ = \sin A$$ Likewise, $$C_2 = \sin B \qquad C_3 = \sin C$$
Moreover, $$D := \left|\begin{array}{ccc} \cos\theta & \sin\theta & - p \\ \cos\phi & \sin\phi & - q \\ \cos\psi & \sin\psi & - r \end{array}\right| = - \left( p C_1 + q C_2 + r C_3 \right) = - \left(\;p \sin A + q \sin B + r \sin C\;\right)$$
Writing $d$ for the circumdiameter of the triangle, the Law of Sines tells us that $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = d$$
Therefore,
$$\begin{align} D &= - \left( \frac{ap}{d} + \frac{bq}{d} + \frac{cr}{d} \right) \\[4pt] &= -\frac{1}{d}\left(\;ap + b q + c r\;\right) \\[4pt] &= -\frac{1}{d}\left(\;2|\triangle COB| + 2|\triangle AOC| + 2|\triangle BOA| \;\right) \\[4pt] &= -\frac{2\;|\triangle ABC|}{d} \end{align}$$
Also, $$C_1 C_2 C_3 = \sin A \sin B \sin C = \frac{a}{d}\frac{b}{d}\sin C= \frac{2\;|\triangle ABC|}{d^2}$$
Finally: $$\frac{D^2}{2C_1C_2C_3} = \frac{4\;|\triangle ABC|^2/d^2}{4\;|\triangle ABC|/d^2} = |\triangle ABC|$$
Yes it is a necessary and sufficient condition if you include as "point of intersection" the point at infinity in the case of parallel lines.
If the two lines $l_1: \; a_1x+b_1y+c_1=0,\quad l_2:\; a_2x+b_2y+c_2=0$ intersect in a point than $l_3=\lambda l_1 +\mu l_2$ also passes though that point. And the their $3 \times 3$ determinant is null.
Viceversa, if the determinant is null, then the rows are dependent and one can be expressed as a linear combination of the others.
-- reply to your comment --
In analytic or affine geometry, the introduction of the homogeneous coordinates render the panorama more ..homogeneous, in fact.
In particular we have the interesting duality points/lines in 2D, and points /planes in 3D lines.
in 2D, two lines always have a common point same as two points have a common line.
Translated to algebra, given
$$l_1: \; a_1x_1+b_1x_2+c_1x_0=0,\quad l_2:\; a_2x_1+b_2x_2+c_2x_0=0$$
then
$$l_1 \cap l_2$$ always has one non-trivial solution, either in terms of a common point $(x_1,x_2,x_0)$ in the above, and in terms of a line common to two points in
$$a x_{1,1}1+bx_{1,2}+cx_{1,0}=0 \; \cap \; a x_{2,1}1+bx_{2,2}+cx_{2,0}=0$$
Then you know that in a homogeneous system prescribing the common point among three lines, a necessary and sufficient condition is that the determinant of the coefficients ($3 \times 3$) be null, because otherwise (rank of the matrix $3$) you have the only solution $(0,0,0)$ which does not correspond to a defined point.
Then if the rank of the matrix is $2$, you have a homogeneous solution $\lambda (\xi, \eta, \zeta)$ which is one point,
and if the rank is $1$ you get a homogeneous solution depending on two parameters, which is a line and which is the same as any of the three in the system.
The same, reverted, for three points sharing a common line.
Best Answer
Basically, the first thing you need to do is find the length of each of the triangle's sides. You can do this by finding the intersection points of the lines, and the corresponding distances between them.
Let the lines be:
$l_1(x)=a_1x+b_1$
$l_2(x)=a_2x+b_2$
$l_3(x)=a_3x+b_3$
Your three points of intersection will then be:
$(\frac{b_2-b_1}{a_1-a_2}, (\frac{b_2-b_1}{a_1-a_2})x+b_1$)
$(\frac{b_3-b_2}{a_2-a_3}, (\frac{b_3-b_2}{a_2-a_3})x+b_2)$
$(\frac{b_1-b_3}{a_3-a_1}, (\frac{b_1-b_3}{a_3-a_1})x+b_3)$
Calculate the distance between each of these points using the formula $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$.
Once you've obtained that, let the three side lengths be $a,b,c$. Using Heron's formula, $s=\frac{a+b+c}{2}:A=\sqrt{s(s-a)(s-b)(s-c)}$.