Find the equation of a line through (1,3) and forming with the axes a triangle of area 6 square units. (Give three solutions).
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Let a and b be the intercept of the line on x and y axis. Then the equation of the line is x/a+x/b = 1
Since this passes through the point P(1,3)
1/a+3/b = 1.<==>b+3a = ab.... (1)
But the area of OAB, where O is (0, 0), A (a,0) and B (0,b) is ab = 12 sq units.
Assume a>0 , b>0. Then ab =12 ...(2).
From (1) and (2) solving for , b+3(12/b) = 12 or
b^2-2b+36 = 0 or (b-6)^2 =0 gives b = 6 and so a =2. So the the required equation is:
x/2+y/6 -1 or 6x+2y -12 = 0 <==> 3x+y -6 =0
Also it is possible that the area of OAB could be (We have to imagine) 12 sq units when a<0 and b>0 or a>0 and b negative.
a<0 or a negative or b is neative but not both. Then ab =-12.
So,
x/(a)+y/(b) = 1 0passes trough (1,3).
So, 1/a +3/b = 1 and ab = -12.
b+3a =-12 and ab =-12.
b+3(-12/b) = -12 or
b^2+12b-36 = 0 or
b = [-12+sqrt(144+144)]/2 or
b = (-6+6sqrt2) = 6(sqrt2-1) or b =-6-6sqrt2 = -6(sqrt2+1)
a =-12/(-6+6sqrt2) = -2(1+sqrt2) or -12/(-6-6sqrt2) = 2(sqrt2 -1)
Therefore the equation of the lines:
-x/2(sqrt2+1) + y/[6(sqrt2-1)] = 1 or
x/[2(sqrt2-1)] -y/[6(sqrt2+1)] = 1
Now we got 3 equations and you can transform them in any desired standard forms like ax+by+c = 0. Or like y = mx+c also.
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