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
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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.
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
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