document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=175673>Question 175673</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A right triangle in the first quadrant is bounded by lines y = 0, y = x, and y = -x+5. Find its area </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #130804 by </B><A HREF=/tutors/aboutme.mpl?userid=gonzo><B>gonzo(654)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=gonzo><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=130804\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the lines y = x and y = -x + 5 are perpendicular to each other since their slopes are negative reciprocals of each other.<BR>\n" );
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document.write( "the line y = 0 means that one of the sides of the triangle will be on the x-axis.<BR>\n" );
document.write( "that line forms the bottom of the triangle.<BR>\n" );
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document.write( "the intersection of the lines y = x, and y = -x + 5, will form the top of the triangle.<BR>\n" );
document.write( "that point of intersection is found by solving y = x, and y = -x + 5 simultaneously.<BR>\n" );
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document.write( "since y = x, you can substitute in the second equation to get x = - x + 5 which solves to be x = 5/2.<BR>\n" );
document.write( "since x = 5/2, then y also equals 5/2, so the intersection point is (5/2,5/2)<BR>\n" );
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document.write( "since the line y = x intersects the line y = 0 at the point (0,0), then (0,0) is also a point on the triangle.<BR>\n" );
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document.write( "since the line y = -x + 5 intersects the line y = 0 at the point (5,0), then (5,0) is the third point of the triangle.<BR>\n" );
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document.write( "if you label the triangle ABC starting from (0,0) and going around clockwise, then the 3 points on the triangle are:<BR>\n" );
document.write( "A (0,0)<BR>\n" );
document.write( "B (5/2,5/2)<BR>\n" );
document.write( "C (5,0)<BR>\n" );
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document.write( "angle B is the right angle because it is the intersects of the two perpendicular lines AB and BC.<BR>\n" );
document.write( "recall that AB is on the line with the equation y = x, and BC is on the line with the equation y = -x+4, and that these lines are perpendicular to each other because the slopes of their equations are negative reciprocals of each other.<BR>\n" );
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document.write( "to find the area, you need to use the formula A = 1/2 base * height.<BR>\n" );
document.write( "since the right angle is at B, the base can be either AB or BC.  we'll use AB.<BR>\n" );
document.write( "since the base is AB, then the height is BC.<BR>\n" );
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document.write( "all we need to do is find the lengths of AB and BC and we can then plug into the equation to get the area.<BR>\n" );
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document.write( "the length of the line segment AB is square root of ((5/2 - 0)^2 + (5/2-0)^2).<BR>\n" );
document.write( "this becomes square root of ((5/2)^2 + (5/2)^2).<BR>\n" );
document.write( "this equals square root of (25/4 + 25/4) equals square root of (50/4) = 3.535533906....<BR>\n" );
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document.write( "the length of the line segment BC is square root of ((5/2-0)^2 + (5/2-5)^2)<BR>\n" );
document.write( "this becomes square root of ((5/2)^2 + (-5/2)^2).<BR>\n" );
document.write( "this equals square root of (25/4 + 25/4) equals square root of (50/4) = 3.535533906....<BR>\n" );
document.write( "---<BR>\n" );
document.write( "area of the triangle is 1/2 bh which equals 1/2 of .....<BR>\n" );
document.write( "square root of (50/4) * square root of (50/4) which equals 1/2 of .....<BR>\n" );
document.write( "(50/4) which equals ...<BR>\n" );
document.write( "50/8 which equals 6.25.<BR>\n" );
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document.write( "area of the triangle is 50/8 = 6.25.<BR>\n" );
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document.write( "a graph of the triangle is shown below.<BR>\n" );
document.write( "look below the graph for further comments.<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=graph+%28800%2C800%2C-2%2C6%2C-2%2C6%2Cx%2C-x%2B5%29 ALIGN=MIDDLE  ALT=\"graph+%28800%2C800%2C-2%2C6%2C-2%2C6%2Cx%2C-x%2B5%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( " </SPAN>\n" );
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