Question

Find the area of a triangle bounded by the y-axis, the line f(x)=12-4x, and the line perpendicular to (f) that passes through the origin.

Answer 1

Answer:

$\dfrac{288}{17}\ un^2.$

Step-by-step explanation:

1. The line y=12-4x has the slope -4, then perpendicular line will have slope $\dfrac{1}{4}.$

The equation of this perpendicular line is

$y-0=\dfrac{1}{4}(x-0),\\ \\4y=x.$

2. The vertices of the triangle are at points:

A: x=0, y=0.

B: x=0, y=12.

C: $\left\{\begin{array}{l}y=12-4x\\4y=x\end{array}\right.\Rightarrow y=12-16y,\ y=\dfrac{12}{17},\ x=\dfrac{48}{17}.$

3. The height of the triangle ABC is x-coordinate of point C, so $h=\dfrac{48}{17},$ the base of the triangle is the length of the segment AB, 12 cm.

4. The area of the triangle ABC is $\dfrac{1}{2}\cdot 12\cdot \dfrac{48}{17}=\dfrac{288}{17}\ un^2.$