document.write( "Question 1123063: One side of an equilateral triangle is along the line 12x+5y-26=0 and the opposite vertex is on the line 12x+5y+13=0. Find the area of the triangle. \n" ); document.write( "
Algebra.Com's Answer #739317 by Boreal(15235)\"\" \"About 
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\n" ); document.write( "5y=-12x+26 or y=-2.4x+5.2
\n" ); document.write( "5y=-12x-13 or y=-2.4x-2.6
\n" ); document.write( "\"graph%28300%2C300%2C-10%2C10%2C-10%2C10%2C-2.4x%2B5.2%2C-2.4x-2.6%2C%285%2F12%29x%2B5.2%29\"
\n" ); document.write( "distance between the lines is the altitude, so pick a point on one line such as (0, 5.2) and find a line perpendicular to that line (negative reciprocal slope product of -1), slope is 5/12, and point is (0, 5.2)\r
\n" ); document.write( "\n" ); document.write( "The equation of the perpendicular line is y-y1=m(x-x1) m slope and (x1, y1) point
\n" ); document.write( "y-5.2=(5/12)x or y=(5/12)x+5.2
\n" ); document.write( "That intersects the other line where the two equations are equal
\n" ); document.write( "(5/12)x+5.2=-(12/5)x-2.6
\n" ); document.write( "(5/12)x+(12/5)x=-7.8
\n" ); document.write( "-7.8=(12/5)x+(5/12)x
\n" ); document.write( "-468=144x+25x, multiplying by 60
\n" ); document.write( "468=-169x
\n" ); document.write( "x=-2.77
\n" ); document.write( "y=4.048
\n" ); document.write( "Want distance between (0, 5.2) and (-2.77, 4.048). That is sqrt (2.77^2+1.152^2)=sqrt(9)=3
\n" ); document.write( "The altitude is 3
\n" ); document.write( "If the side is s, the altitude is (s/2)*sqrt(3) and area is (s^2/4)*sqrt(3)
\n" ); document.write( "Altitude is 3, so half of a side is 3/sqrt(3) or sqrt(3) and side is 2 sqrt(3)
\n" ); document.write( "Area is therefore [2 sqrt(3)^2]/4* sqrt(3), or 3 sqrt (3) units
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