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

Answer:

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

Step-by-step explanation:

we have

$f(x)=4-\frac{5}{7}x$

The slope of the given linear function is

$m=-\frac{5}{7}$

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

$m_1=-\frac{5}{7}$

substitute

$(-\frac{5}{7})*m_2=-1$

$m_2=\frac{7}{5}$

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

$y=\frac{7}{5}x$

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

$A=\frac{1}{2}(AB)(BC)$

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

$A=\frac{1}{2}(3.255)(2.325)$

$A=3.78\ units^2$