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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.

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You don't have to sketch a graph (though in this case it certainly makes it easier to solve the rest of the question). You can take two of the lines and solve the equations simultaneously to find the $x$ and $y$ coordinates of a vertex of the triangle; repeat for the other two possible pairs of lines. – Rahul Aug 16 '12 at 20:24

Following rschwieb's advice:

Note that

$\ \ \ \bullet$ $\color{darkgreen}{x=0}$ is the line coinciding with the $y$-axis.

$\ \ \ \bullet$ $\color{darkblue}{x+y=0}$ has standard form $y=-x$; so, its graph is the line of slope $-1$ passing through the origin.

$\ \ \ \bullet$ $\color{maroon}{y=x-1}$ is the equation of the line of slope $1$ with $y$-intercept $(0,-1)$.

The graphs of these lines are shown below:

enter image description here

The triangle in question is shaded light green (on my display) above. To find the coordinates of the vertices of this triangle, you have to solve systems of equations.

Apparently there is a vertex around the point $(1/2,-1/2)$. Let's check that. Here, we need to find the intersection of the lines $\color{darkblue}{x+y=0}$ and $\color{maroon}{y=x-1}$. I'll leave it to you to solve this system. It indeed has the solution $(1/2,-1/2)$.

Once you have the coordinates of all three vertices, computing the area of the triangle should be easy. For instance, you can take the base to be $1$ (along the $y$-axis) and the corresponding height to be $1/2$.

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

  1. 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.

  2. 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 $\{f_1, f_2 \}, \ \{f_1, f_3 \}, \ \{f_2, f_3 \}$.

  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].

  1. 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

$S(ABC)=\frac{1}{2}|0 \cdot (-\frac{1}{2}) + \frac{1}{2} \cdot (-1) + 0 \cdot 0 - \frac{1}{2} \cdot 0 - 0 \cdot (-\frac{1}{2}) - 0 \cdot (-1) |=\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$.

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$x+y=0$, and $x=0$ and $y=x-1$ are the equations of three separate lines. To make any headway you will have to graph these lines.

I've heard this "could be anywhere" thing from students, but it's important to know that (for nonhorizontal lines) each $y$ coordinate will correspond to just one $x$ coordinate, and then that pair determines a single point on the line.

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