# Find the equation of the line of slope -3/4 that forms with the coordinate axes a triangle which has the area of 24 square units.

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First the form of the equation would be;

y= mx + b, where m is the slop.

==> y = (-3/4)x + b

when x = 0, then y = (-3/4) 0 + b= b

then the point where the line intersect with the y-axis is (0,b)

when y = 0 , then (-3/4) x +b= 0

==> (3/4) x= b

==> x= (4/3)b

then the point where the line intersect with the x-axis is (4b/3, 0)

but, the area of the triangle is:

A= 1/2 * 4b/3 * b

24 = 1/2 * 4b/3 *b

24 = 1/2 * 4b^2/3 = 4b^2 /6

==> b^2 = 24*6/4

==> b^2 = 36

==> b = 6, -6

then the eqution of teh slope is:

y = (-3/4) x + 6 and y= (-3/4)x -6

The equation of the line, whose slope is known, is:

y = mx+b, where m=-3/4 (from enunciation).

If we set y=0, the equation will be:

0 = (-3/4)x+b

When y=0, the line is intercepting ox axis.

(3/4)x = b

x = (4/3)b

The area of the triangle is A = (1/2)a*b = (1/2)*(4/3)b*b.

But we know, from enunciation that the area is 24 square units.

24 = (4/6)*b^2

b^2=36

b1=6 and b2=-6

The equations of the line which has the slope m=-3/4 are:

y = (-3/4)x+b1 and y = (-3/4)x+b2

**y = (-3/4)x+6 and y = (-3/4)x-6**

Let y = mx +c be the equation of the line.As the slope is -3/4,

the could be y = (-3/4)x+c.

The x and y intercepts are got by putting y= 0 and solving for x. So 0 = -3/4(x)+c. x = -c /(-3/4) = 4c/3.

And y intercept is y = c.

So the area of the triangle formed by the x intercept y intercept and the line y = (-3/4)x+c is

(1/2)( xintercept)(y intercept) = (1/2)(4c/3)(c) = 2c^2/3. But this is given to be equal to 24 .So.

2c^2/3 = 24 Or

c^2 = 24*3/2 = 36. Or

c = sqrt36 = = 6 or c = -6.

So the equation of the line is:

y = (-3/4)x+6 Or

y = (-3/4)x-6.