Question

Find the area of a triangle bounded by the y-axis, the line f(x)=9−2/3x, and the line perpendicular to f(x) that passes through the origin. Area =

Answer 1

Answer:

The area of the triangle is 18.70 sq.units.

Step-by-step explanation:

It is provided that a triangle is bounded by the y-axis, the line $f(x)=y=9-\frac{2}{3}x$.

The slope of the line is: $m_{1}=-\frac{2}{3}$

A perpendicular line passes through the origin to the line f (x).

The slope of this perpendicular line is:$m_{2}=-\frac{1}{m_{1}}=\frac{3}{2}$

The equation of perpendicular line passing through origin is:

$y=\frac{3}{2}x$

Compute the intersecting point between the lines as follows:

$y=9-\frac{2}{3}x\\\\\frac{3}{2}x=9-\frac{2}{3}x\\\\\frac{3}{2}x+\frac{2}{3}x=9\\\\\frac{13}{6}x=9\\\\x=\frac{54}{13}$

The value of y is:

$y=\frac{3}{2}x=\frac{3}{2}\times\frac{54}{13}=\frac{81}{13}$

The intersecting point is $(\frac{54}{13},\ \frac{81}{13})$.

The y-intercept of the line f (x) is, 9, i.e. the point is (0, 9).

So, the triangle is  bounded by the points:

(0, 0), (0, 9) and $(\frac{54}{13},\ \frac{81}{13})$

Consider the diagram attached.

Compute the area of the triangle as follows:

$\text{Area}=\frac{1}{2}\times 9\times \frac{54}{13}=18.69231\approx 18.70$

Thus, the area of the triangle is 18.70 sq.units.