Find the area of a triangle bounded by the y axis, the line f(x)=5−56x, and the line perpendicular to f(x) that passes through the origin. Area = ?

Answer 1

y = 5-56x

y=x/56 is the perpendicular line through the origin, it's nearly flat nearly y=0

y axis is the line x=0

the 3 lines intersect at points A(0,5), B(0,0) and C(68/627, 68/3512) or about C(.11,.02)

the 3 points are vertices of a right triangle with sides AC and BC and hypotenuse AB. Area = AC(BC)/2 = one half of the product of the two shorter sides

AC^2 = (68/627)^2 + (5-68/3512)^2

BC = (68/627)^2 + (68/3512)^2

AC = about sqr(.11^2 + 4.8^2) = about 4.8

BC = about sqr(.0125) = 0.112

Area = about (4.8)(.112)/2 = 0.267

The area is a less more than a right triangle of height 5 and base = 1/9

5(1/9)/2 = 5/18 = about 0.28

or use the base as 68/627

bh/2 = (68/627)(5)/2 = 0.27