Question

Find the area of a triangle bounded by the y-axis, the line f(x) = 4−5/7x, and the line perpendicular to f(x) that passes through the origin. (Round your answer to two decimal places.)

Answer 1

The area of triangle is
`3.78\ units^2`

Explanation:

we have

`f(x)=4-(5)/(7)x`

The slope of the given linear function is

`m=-(5)/(7)`

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

`m_1*m_2=-1`

Find the slope
`m_2` of the line perpendicular to the given linear function

we have

`m_1=-(5)/(7)`

substitute

`(-(5)/(7))*m_2=-1`

`m_2=(7)/(5)`

Find the equation of the line perpendicular to the given linear function that passes through the origin

The line represent a direct variation, because the line passes through the origin

The equation is

`y=(7)/(5)x`

Find the area of triangle bounded by the y-axis, the line f(x) = 4−5/7x, and the line perpendicular to f(x) that passes through the origin

using a graphing tool

see the attached figure

The vertices of the triangle are

A(0,0),B(1.892,2.649),C((0,4)

The area of the right triangle ABC is

`A=(1)/(2)(AB)(BC)`

the formula to calculate the distance between two points is equal to

`d=\sqrt{(y2-y1)^(2)+(x2-x1)^(2)}`

Find the distance AB

`d_A_B=\sqrt{(2.649-0)^(2)+(1.892-0)^(2)}`

`d_A_B=3.255\ units`

Find the distance BC

`d_B_C=\sqrt{(4-2.649)^(2)+(0-1.892)^(2)}`

`d_B_C=2.325\ units`

Find the area of the right triangle ABC

`A=(1)/(2)(3.255)(2.325)`

`A=3.78\ units^2`