Find area of a triangle given the equation of sides Find the coordinates of the vertices of the triangle $ABC$ formed by the intersection of the lines $x + y = 0$, $x = 0$ and $y = x - 1$. Hence, find the area of the triangle.
I tried to sketch a graph but since the $y$ coordinate could be anywhere, I am having trouble working it out. 

As student to student. If I had to solve such problem in my school, generally, I would have done it the following steps:


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*We have 3 equations $f_i$ of separate lines which are $a_i\cdot x + b_i\cdot y = c_i$ where $a_i,b_i,c_i$ given numbers.

*First we should find the coordinates of each vertex of the triangle by solving system of 2 equations. We have 3 equations, so we have 3 different systems $\{f_1, f_2 \}, \ \{f_1, f_3 \}, \ \{f_2, f_3 \}$.

*If we have a system $ \{ a\cdot x + b \cdot y = c; \  d\cdot x + e \cdot y = f\}$ its solution is $x=\frac{b\cdot f - c \cdot e}{b\cdot d - a \cdot e} $; and  $y=\frac{c\cdot d - a \cdot f}{b\cdot d - a \cdot e}.$
In your case we have 3 equations $x+y=0; \ x=0; \ -x+y=-1. $
i. First system are $\{x+y=0; \ x=0\}$ here a=1, b=1, c=0, d = 1, e=0, f=0. It give us $x=\frac{b\cdot f - c \cdot e}{b\cdot d - a \cdot e}=\frac{1\cdot 0 - 0 \cdot 0}{1\cdot 1 - 1 \cdot 0}=0$ and $y=\frac{c\cdot d - a \cdot f}{b\cdot d - a \cdot e}=\frac{0\cdot 1 - 1 \cdot 0}{1\cdot 1 - 1 \cdot 0}=0$ - coordinates of the first vertex A. So $A[0,0]$.
ii. Second system are $\{x+y=0; \ -x+y=-1 \}$ here a=1, b=1, c=0, d = -1, e=1, f=-1. It give us $x=\frac{1}{2}$ and $y=-\frac{1}{2}$ - coordinates of the second vertex B. So $B[\frac{1}{2},-\frac{1}{2}]$.
iii. Third system are $\{x=0; \ -x+y=-1 \}$ here a=1, b=0, c=0, d = -1, e=1, f=-1. It give us $x=0$ and $y=-1$ - coordinates of the third vertex C. So C[0,-1].


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*If we have triangle ABC with $A[x_1,y_1], \ B[x_2,y_2], C[x_3,y_3]$ then its area $S(ABC)=\frac{1}{2}|x_1 \cdot y_2 + x_2 \cdot y_3 + x_3 \cdot y_1 - x_2 \cdot y_1 - x_3 \cdot y_2 - x_1 \cdot y_3 |$.


In your case $A[0,0], B[\frac{1}{2},-\frac{1}{2}], C[0,-1] $ so $x_1=0, y_1=0, x_2=\frac{1}{2}, y_2=-\frac{1}{2}, x_3=0, y_3=-1$. So
$S(ABC)=\frac{1}{2}|0 \cdot (-\frac{1}{2}) + \frac{1}{2} \cdot (-1) + 0 \cdot 0 - \frac{1}{2} \cdot 0 - 0 \cdot (-\frac{1}{2}) - 0 \cdot (-1) |=\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$.
