Question

Find the area of a triangle bounded by the y-axis, the line f(x)=9-6/7x, and the line perpendicular to f(x) that passes through the origin.

Answer 1

Answer:

  20 1/85 ≈ 20.01 square units

Step-by-step explanation:

There are several ways we can find the area of the shaded triangle in the attached figure. We can use proportions, and we can use the triangle area formula.

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proportions

We know all of the triangles involved are right triangles, and that AB, AD, and BD are their hypotenuses.

The location of point B is the y-intercept of the boundary line: y = 9.

The location of point D is the x-intercept of the equation of the boundary line, so will be found where y=0:

  y = 9 -6/7x

  0 = 9 -6/7x

  0 = 10.5 -x . . . . . multiply by 7/6

  x = 10.5

That means the scale factor between ΔABC and ΔDAC is ...

  AB/DA = 9/10.5 = 6/7

The ratio of the triangle areas is the square of the scale factor:

  (area ΔABC)/(area ΔDAC) = (6/7)² = 36/49

The ratio of the area of ΔABC to the total of the areas ΔABC+ΔDAC will then be ...

  area ΔABC/(area ΔABD) = 36/(36+49) = 36/85

The formula for the area of a triangle gives the area of ΔABD:

  A = 1/2bh = 1/2(10.5)(9) = 47.25 . . . . square units

Then the area of ΔABC is ...

  area ΔABC = (fraction of whole) × (whole)

  = (36/85)×47.25 = 20 1/85 ≈ 20.01 . . . . square units

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from triangle dimensions

To use the area formula for ΔABC, we need to know a base and height. We already know the length of side AB = 9, so we only need to find the x-coordinate of point C. That is at the intersection of the boundary lines.

The perpendicular line AC has a slope that is the opposite reciprocal of the slope of line BC: -1/(-6/7) = 7/6. Since line AC goes through the origin, its equation is ...

  y = 7/6x

Substituting for y in the equation y = f(x), we have ...

  7/6x = 9 -6/7x

  (7/6 +6/7)x = 9 . . . . . add 6/7x

  85/42x = 9   ⇒   x = 9(42/85) = 4 38/85

This is the height from AB to C of ΔABC. So, the area of ΔABC is ...

  area = 1/2bh = 1/2(9)(4 38/85) = 20 1/85 . . . . square units

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Additional comment

We could finish the solution of the coordinates of C, then find AC and BC using the distance formula. These lengths could then be used in the area formula for a triangle to find area ΔABC. We suspect either of the methods used here involves less work and opportunity for mistakes.