Question

Find the area of a triangle bounded by the y-axis, the line

Answer 1

The area is 68.47.

A drawing of the triangle is attached.

We first plot the y-intercept of the line given to us, f(x)=34-4x.  This is in the form f(x)=mx+b, so the y-intercept, b, is 34.  

Next we use the slope to find the next points of the line.  The slope is -4, which can be written as -4/1 (rise/run).  This means from the y-intercept, we go down 4 and over 1 to plot the next point, and continue in this manner.  We now have the y-axis and this line drawn.

Next we use the slope of the line to find the slope of the perpendicular line described.  Two lines are perpendicular if their slopes are negative reciprocals of each other.  The slope of the first line was -4/1; this means the slope of this line must be 1/4.  We also know this point goes through the origin, so 0 is the y-intercept, and we have the equation f(x)=1/4x+0.

To graph this, plot a point at 0, the y-intercept.  The slope is 1/4, so we go up 1 and over 4 to the next point.  This happens to be on the line we previously drew.

To find the area of this triangle, we need to identify the base and height.  The height will be perpendicular to the base.  The two sides that form a right angle are the bottom and the right hand side, so the bottom is the base and the right hand side is the height.

Now we find the length of each of these sides using the distance formula.  For the height:

$d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}=\sqrt{(34-1)^2+(0-4)^2} \\ \\=\sqrt{33^2+(-4)^2}=\sqrt{1089+16}=\sqrt{1105}\approx33.24$

To find the length of the base:
$d=\sqrt{(0-1)^2+(0-4)^2}=\sqrt{(-1)^2+(-4)^2}=\sqrt{1+16}=\sqrt{17} \\ \\\approx4.12$

The area of a triangle is given by A=1/2bh:
A=1/2(33.24)(4.12) = 68.4744 ≈ 68.47