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.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! rewrite as
5y=-12x+26 or y=-2.4x+5.2
5y=-12x-13 or y=-2.4x-2.6
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)
The equation of the perpendicular line is y-y1=m(x-x1) m slope and (x1, y1) point
y-5.2=(5/12)x or y=(5/12)x+5.2
That intersects the other line where the two equations are equal
(5/12)x+5.2=-(12/5)x-2.6
(5/12)x+(12/5)x=-7.8
-7.8=(12/5)x+(5/12)x
-468=144x+25x, multiplying by 60
468=-169x
x=-2.77
y=4.048
Want distance between (0, 5.2) and (-2.77, 4.048). That is sqrt (2.77^2+1.152^2)=sqrt(9)=3
The altitude is 3
If the side is s, the altitude is (s/2)*sqrt(3) and area is (s^2/4)*sqrt(3)
Altitude is 3, so half of a side is 3/sqrt(3) or sqrt(3) and side is 2 sqrt(3)
Area is therefore [2 sqrt(3)^2]/4* sqrt(3), or 3 sqrt (3) units
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