Question

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

Answer 1

The area of a triangle bounded by the y-axis is 8.49 square units

Solution:

Given that f(x) =
`6 - (5)/(7)x`

`\text { Let } y=6-(5)/(7) x`

`y = (-5)/(7)x + 6`

On comparing the above equation with slope intercept form.i.e

y = mx + c

where "m" is the slope and "c" is the y-intercept

So slope =
`(-5)/(7)`

We know product of slopes of perpendicular line and given line is always -1

Slope of perpendicular line is given as:

`= (7)/(5)`

Equation of perpendicular line passing through origin (0, 0) is:

y = mx + c

`y = (7)/(5)x + 0\\\\y = (7)/(5)x`

Intersecting point between the lines is:

`(7)/(5)x = 6 - (5)/(7)x\\\\(7)/(5)x + (5)/(7)x = 6`

`(74x)/(35) = 6\\\\x = 2.83`

We know that
`y = (7)/(5)x`

`y = (7 * 2.83)/(5)\\\\y = 3.962`

Point is (2.83, 3.962)

y intercept of line is
`y = 6 - (5)/(7)x\\`

Put x = 0

Therefore y = 6

So the triangle is bounded by the points (0, 0) and (0, 6) and (2.83, 3.962)

`\text { Area of triangle }=(1)/(2) * 6 * 2.83=8.49`

Thus area of triangle is 8.49 square units