# Find the equations of the lines which have the slope -3/4 and form with the coordinate axes a triangle of area 2 sq. units

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We are given that the slope of the line is `m=-3/4` , and that the area of the triangle formed by the line and the coordinate axes is 2.

Let the x-intercept be x and the y intercept be y.

Since the slope is `-3/4` we have `y/(-x)=(-3/4)==>y=3/4 x`

Since the area of the triangle is 2 we have `A=2=1/2xy ==> y=4/x`

`4/x=3/4 x`

`x^2=16/3`

`x=+-sqrt(16/3)=+-4/sqrt(3) = +-(4sqrt(3))/3`

Then `y=+-sqrt(3)`

(1)IfÂ `x=(4sqrt(3))/3,y=sqrt(3)` are the intercepts the line is `y=-3/4x+sqrt(3)`

Note that the area is `1/2((4sqrt(3))/3)(sqrt(3))=2`

(2) If `x=(-4sqrt(3))/3,y=-sqrt(3)` are the intercepts the line is `y=-3/4x-sqrt(3)`

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The lines are `y=-3/4x+sqrt(3),y=-3/4x-sqrt(3)`

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Thanks :)