Question 1016825
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show that the points (4,0), (2,1), (-1,-5) are the vertices of a right triangle and find its area.
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This is your triangle.


<TABLE>
  <TR>
  <TD>
{{{drawing( 250, 250,  -5, 5, -5, 5, 
            grid(1),
            line( 4,  0,  2,  1), 
            line( 4,  0, -1, -5),
            line( 2,  1, -1, -5)
)}}}
 </TD>
 </TR>
</TABLE>

<pre>
Calculate the distances between the given points (4,0), (2,1), (-1,-5).

1. Points (4,0) and (2,1). The distance a = {{{sqrt((4-2)^2 + (0-1)^2)}}} = {{{sqrt(2^2 + 1^2)}}} = {{{sqrt(5)}}}.

2. Points (2,1) and (-1,-5). The distance b = {{{sqrt((2-(-1))^2 + (1-(-5))^2)}}} = {{{sqrt(3^2 + 6^2)}}} = {{{sqrt(9 + 36)}}} = {{{sqrt(45)}}}.

3. Points (4,0) and (-1,-5). The distance c = {{{sqrt((4-(-1))^2 + (0-5)^2)}}} = {{{sqrt(5^2 + 5^2)}}} = {{{sqrt(25 + 25)}}} = {{{sqrt(50)}}}.

Now you see that {{{a^2}}} + {{{b^2}}} = 5 + 45 = 50 = {{{c^2}}}.

It is just enough to conclude that the triangle is right-angled.

Next, the area of the triangle is {{{(a*b)/2}}} = {{{(sqrt(5)*sqrt(45))/2}}} = {{{sqrt(225)/2}}} = {{{15/2}}} = 7.5 {{{cm^2}}}.
</pre>