Question 1099314: Determine graphically,the co-ordinates of the vertices of th triangle bounded by 2x-3y+6=0;2x+3y-18=0;y-2=0 Also find the area of this triangle
Found 2 solutions by Alan3354, KMST:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Determine graphically,the co-ordinates of the vertices of th triangle bounded by 2x-3y+6=0;2x+3y-18=0;y-2=0 Also find the area of this triangle
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Graph the 3 equations.
The intersections are the 3 vertices.
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I would find the area using numbers.
Answer by KMST(5328)  (Show Source):
You can put this solution on YOUR website! Determining graphically means we have to start by graphing.
 <--->  is easy to graph:
 .
The other two lines can be esily graphed using their x- and y-intercepts.
Once the intercepts for a line are found,
we just plot the points, and draw a line through them.
 <--> 
has x- and y-intercepts (-3,0) and (0,2) respectively,
that can be found by susbtituting 0 for y and x respectively.
The x- and y-intercepts for
 <--> 
can be similarly found to be (9,0) and (0,6) .
 .
From the graph it looks like the vertices of the triangle are (0,2) , (3,4) , and (6,2) .
With the graph, it is easy to see that triangle as one with
a horizontal base of length  ,
and a vertical height  .
Then, the area of the triangle can be calculated as  .
DO WE NEED TO REMEMBER THE FORMULA FOR THE AREA OF A TRIANGLE, AND DO THE CALCULATION?
Not really. It is obvious that the triangle is half of the rectangle outlined below, so we could just count squares: one rectangle = 12 squares, one hald rectangle = 6 squares. .
WHAT IF THE TEACHER WANTS YOU TO SHOW MORE WORK?
I see two places where you would have to show that you really understand how to figure out the problem:
1) Any side of a triangle (a segment between two vertices) can be called the base, and then the perpendicular distance to the other vertex is the height.
In this case it is easiest to call the horizontal segment between (0,2) and (6,2) the base of the triangle,
and then the vertical distance from that base to point (3,4) would be the height.
Theoretically you would calculate the length of the horizontal base as the difference in x-coordinates of the endpoints of that base, so  .
Theoretically you would calculate the vertical height as the difference in y-coordinates between the base, and the other vertex, so  .
2) Would your teacher insist that you verify the coordinates with some calculations?
Maybe we should really verify that the lines intersect at points with exactly those coordinates,
by substituting those coordinates into the equations of the lines,
because sometimes things are not quite exactly the way they look.
We do not need to verify (0,2) ,
which we have shown above that is a point that
is part of blue line , and
is part of red line .
Substituting  , the coordinates of (3,4),
we see that it is indeed a solution of  .
Substituting  , the coordinates of (6,2),
we see that it is indeed a solution of  .
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