Question

Find the area of a triangle bounded by the y-axis, the line f(x)=9-\frac{6}{7}x, and the line perpendicular to f(x) that passes through the origin. Type your answer rounded to the nearest hundredth.The area is Answer square units.

Answer 1

To understand the question let us draw a figure

We need to find the area of triangle OAB

Its base is OA with a length of 9

Its height BC

We need to find BC

It is equal to the x-coordinate of point B

Then we have to solve the equations of the 2 perpendicular lines BA and BO to find the x coordinate of B

The equation of BA is given

$y=9-\frac{6}{7}x\rightarrow(1)$

Since the slope of perpendicular lines are opposite reciprocal of each other

Since the slope of line AB is -6/7, then the slope of OB is 7/6

Since the line OB passes through the origin, then its y-intercept = 0

Then the equation of BO is

$y=\frac{7}{6}x\rightarrow(2)$

Equate (1) and (2)

$\frac{7}{6}x=9-\frac{6}{7}x$

Add 6/7 x to both sides

$\begin{gathered} \frac{7}{6}x+\frac{6}{7}x=9-\frac{6}{7}x+\frac{6}{7}x \\ \frac{85}{42}x=9 \end{gathered}$

Divide both sides by 85/42

$x=\frac{378}{85}$

Then the height of the triangle is 378/85

Then the area of the triangle is

$\begin{gathered} A=\frac{1}{2}(9)(\frac{378}{85}) \\ A=20.01176471 \end{gathered}$

Then the area is 20.01 square units to the nearest hundredth