document.write( "Question 715519: Find the area of the triangle formed by the (x,y)-axes and the line with equation 4x+3y-12=0 \n" ); document.write( "

Algebra.Com's Answer #439538 by KMST(5328) $\"\"$ $\"About$

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The x- and y-intercepts of the line with equation
\n" ); document.write( " $\"4x%2B3y-12=0\"$ <--> $\"4x%2B3y=12\"$ are easy to find.
\n" ); document.write( "For $\"x=0\"$ --> $\"4%2A0%2B3y=12\"$ --> $\"0%2B3y=12\"$ --> $\"3y=12\"$ --> $\"y=4\"$
\n" ); document.write( "gives you point (0,4) as the y-intercept.
\n" ); document.write( "For $\"y=0\"$ --> $\"4x%2B3%2A0=12\"$ --> $\"4x%2B0=12\"$ --> $\"4x=12\"$ --> $\"x=3\"$
\n" ); document.write( "gives you point (3,0) as the x-intercept.
\n" ); document.write( "Those 2 points and point (0,0) , the origin, are the vertices of your triangle:
\n" ); document.write( "

\n" ); document.write( "You could say that the length of the base of your triangle is $\"b=3\"$ units
\n" ); document.write( "and the height is $\"h=4\"$ units,
\n" ); document.write( "so the area (in square units) is
\n" ); document.write( " $\"b%2Ah%2F2=3%2A4%2F2=highlight%286%29\"$ \n" ); document.write( "

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