<?xml version="1.0" encoding="UTF-8"?><rss version="2.0"
	xmlns:content="http://purl.org/rss/1.0/modules/content/"
	xmlns:dc="http://purl.org/dc/elements/1.1/"
	xmlns:atom="http://www.w3.org/2005/Atom"
	xmlns:sy="http://purl.org/rss/1.0/modules/syndication/"
		>
<channel>
	<title>Comments on: Finding the area of a triangle based on vertices lines?</title>
	<atom:link href="http://www.thearealist.com/triangle-area/finding-the-area-of-a-triangle-based-on-vertices-lines/feed" rel="self" type="application/rss+xml" />
	<link>http://www.thearealist.com/triangle-area/finding-the-area-of-a-triangle-based-on-vertices-lines</link>
	<description></description>
	<lastBuildDate>Fri, 03 Feb 2012 13:07:59 -0500</lastBuildDate>
	<generator>http://wordpress.org/?v=2.8.4</generator>
	<sy:updatePeriod>hourly</sy:updatePeriod>
	<sy:updateFrequency>1</sy:updateFrequency>
		<item>
		<title>By: Brian</title>
		<link>http://www.thearealist.com/triangle-area/finding-the-area-of-a-triangle-based-on-vertices-lines/comment-page-1#comment-1367</link>
		<dc:creator>Brian</dc:creator>
		<pubDate>Mon, 30 Nov 2009 13:15:59 +0000</pubDate>
		<guid isPermaLink="false">http://www.thearealist.com/triangle-area/finding-the-area-of-a-triangle-based-on-vertices-lines#comment-1367</guid>
		<description>-3x+y=4, and 2x+y=4 intersect at (0,4), call that point A
x-2y=2 and 2x+y=4 intersect at (2,0), call that point B
x-2y=2 and -3x+y=4 intersect at (-2,-2), call that point C
Side AB is on line 2x+y=4, which has slope -2
Side BC is on line x-2y=2, which has slope 1/2
AB and BC are perpendictular (-2 * 1/2 = -1)
The length of AB is sqrt[20]
The length of BC is sqrt[20]
The area of ABC is sqrt[20] * sqrt[20] / 2 = 10&lt;br&gt;&lt;b&gt;References : &lt;/b&gt;&lt;br&gt;</description>
		<content:encoded><![CDATA[<p>-3x+y=4, and 2x+y=4 intersect at (0,4), call that point A<br />
x-2y=2 and 2x+y=4 intersect at (2,0), call that point B<br />
x-2y=2 and -3x+y=4 intersect at (-2,-2), call that point C<br />
Side AB is on line 2x+y=4, which has slope -2<br />
Side BC is on line x-2y=2, which has slope 1/2<br />
AB and BC are perpendictular (-2 * 1/2 = -1)<br />
The length of AB is sqrt[20]<br />
The length of BC is sqrt[20]<br />
The area of ABC is sqrt[20] * sqrt[20] / 2 = 10<br /><b>References : </b></p>
]]></content:encoded>
	</item>
	<item>
		<title>By: sahsjing</title>
		<link>http://www.thearealist.com/triangle-area/finding-the-area-of-a-triangle-based-on-vertices-lines/comment-page-1#comment-1366</link>
		<dc:creator>sahsjing</dc:creator>
		<pubDate>Mon, 30 Nov 2009 12:25:59 +0000</pubDate>
		<guid isPermaLink="false">http://www.thearealist.com/triangle-area/finding-the-area-of-a-triangle-based-on-vertices-lines#comment-1366</guid>
		<description>Draw all three lines on x-y plane, and find all intersection points: (0,4), (2,0) and (-2,-2).
Since the triangle can be split into two triangles with a common base = 2+4/3 = 10/3, the area is equal to
(1/2)(10/3)(4+2) = 10 units^2
---------
Attn: Try to find to simpler way to get it done.&lt;br&gt;&lt;b&gt;References : &lt;/b&gt;&lt;br&gt;</description>
		<content:encoded><![CDATA[<p>Draw all three lines on x-y plane, and find all intersection points: (0,4), (2,0) and (-2,-2).<br />
Since the triangle can be split into two triangles with a common base = 2+4/3 = 10/3, the area is equal to<br />
(1/2)(10/3)(4+2) = 10 units^2<br />
&#8212;&#8212;&#8212;<br />
Attn: Try to find to simpler way to get it done.<br /><b>References : </b></p>
]]></content:encoded>
	</item>
</channel>
</rss>

