Q:

Find the area of a triangle bounded by the y-axis, the line f(x) = 6 − (5/7)x, and the line perpendicular to f(x) that passes through the origin. (Round your answer to two decimal places.)Edit: I got the answer! It's 8.51

Accepted Solution

A:
The area of a triangle bounded by the y-axis is 8.49 square unitsSolution:Given that f(x) = [tex]6 - \frac{5}{7}x[/tex][tex]\text { Let } y=6-\frac{5}{7} x[/tex][tex]y = \frac{-5}{7}x + 6[/tex]On comparing the above equation with slope intercept form.i.e y = mx + c where "m" is the slope and "c" is the y-interceptSo slope = [tex]\frac{-5}{7}[/tex]We know product of slopes of perpendicular line and given line is always -1Slope of perpendicular line is given as:[tex]= \frac{7}{5}[/tex]Equation of perpendicular line passing through origin (0, 0) is:y = mx + c[tex]y = \frac{7}{5}x + 0\\\\y = \frac{7}{5}x[/tex]Intersecting point between the lines is:[tex]\frac{7}{5}x = 6 - \frac{5}{7}x\\\\\frac{7}{5}x + \frac{5}{7}x = 6[/tex][tex]\frac{74x}{35} = 6\\\\x = 2.83[/tex]We know that [tex]y = \frac{7}{5}x[/tex][tex]y = \frac{7 \times 2.83}{5}\\\\y = 3.962[/tex]Point is (2.83, 3.962)y intercept of line is [tex]y = 6 - \frac{5}{7}x\\[/tex]Put x = 0Therefore y = 6So the triangle is  bounded by the points (0, 0) and (0, 6) and (2.83, 3.962)[tex]\text { Area of triangle }=\frac{1}{2} \times 6 \times 2.83=8.49[/tex]Thus area of triangle is 8.49 square units