calculus question need help

Question

Find the area of a triangle bounded by the  axis, the line , and the line perpendicular to  that passes through the origin.

Answer 1

Slope of given line is - 3/4 and because slope of perpendicular line is 4/3 and equation y = (4/3)x.

Let find x-coordinate of intersection point: (4/3)x = 9 - (3/4)x; 16x = 108 - 9x; 25x = 108, x = 108/25 = 4.32.

Area A = ∫₀^4.32((9 - 3x/4 - 4x/3)dx = (9x - 25x²/24)₀^4.32 = 19.44

Answer 2

Video answer did not complete, so here is a condensed answer:

1. The line ⊥ to the line with equation y = 9 - 3/4 (x) has slope 4/3 (negative reciprocal).
2. Therefore the equation of the ⊥ is y = 4/3 (x), since the y-intercept is at (0,0).
3. Using the substitution method find the point of intersection of those two lines: P(x₁, y₁).
4. The line y = 0 intersects y = 9 - 3/4 (x) at Q(0,9) and the line y = 4/3 (x) at R(0,0).
5. So points (0,9), (0,0) , (x₁, y₁) are the vertices of a right triangle with right angle at P.
6. Find the distance from P to Q (PQ) using distance formula.
7. Find the distance from P to R (PR) using the distance formula.
8. Those are the legs of a right triangle and can be used as the base and height in the formula for the area of a triangle.
9. So the answer can be found by A = 1/2 (PR)(PQ).

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