# 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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