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3 years ago

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Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s). Line A passes t

hrough points (-1, 5) and (3, -1). Line B passes through points (7, 2) and (6, -1). At what point does line A intersect line B?

1 answer:

natulia [17]3 years ago

5 0

Line A
slope = (-1 - 5)/(3+1) = -6/4 = -3/2
y = mx + b
5 = -3/2(-1) + b
b = 5 - 3/2
b = 7/2
equation of line A: y = -3/2x + 7/2

Line B
slope = (-1 - 2)/(6-7) = -3/-1 = 3
y = mx + b
2 = 3(7) + b
b = 2- 21
b = - 19
equation of line B: y = 3x -19

line A intersect line B when -3/2x + 7/2 = 3x -19
so
-3/2x + 7/2 = 3x -19
3x + 3/2x = 19 + 7/2
9/2x = 45/2
x = (45/2) * (2/9)
x = 5

y = 3x - 19 = 3(5) -19 = 15 -19 = -4

(5 , -4)

answer
line A intersect line B at (5, -4)

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Since you mentioned calculus, perhaps you're supposed to find the area by integration.

The square is circumscribed by a circle of radius 6, so its diagonal (equal to the diameter) has length 12. The lengths of a square's side and its diagonal occur in a ratio of 1 to sqrt(2), so the square has side length 6sqrt(2). This means its sides occur on the lines $x=\pm3\sqrt2$ and $y=\pm3\sqrt2$ .

Let $R$ be the region bounded by the line $x=3\sqrt2$ and the circle $x^2+y^2=36$ (the rightmost blue region). The right side of the circle can be expressed in terms of $x$ as a function of $y$ :

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Then the area of this circular segment is

$\displaystyle\iint_R\mathrm dA=\int_{-3\sqrt2}^{3\sqrt2}\int_{3\sqrt2}^{\sqrt{36-y^2}}\,\mathrm dx\,\mathrm dy$

$=\displaystyle\int_{-3\sqrt2}^{3\sqrt2}(\sqrt{36-y^2}-3\sqrt2)\,\mathrm dy$

Substitute $y=6\sin t$ , so that $\mathrm dy=6\cos t\,\mathrm dt$

$=\displaystyle\int_{-\pi/4}^{\pi/4}6\cos t(\sqrt{36-(6\sin t)^2}-3\sqrt2)\,\mathrm dt$

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Alternatively, you can compute the area of $R$ in polar coordinates. The line $x=3\sqrt2$ becomes $r=3\sqrt2\sec\theta$ , while the circle is given by $r=6$ . The two curves intersect at $\theta=\pm\dfrac\pi4$ , so that

$\displaystyle\iint_R\mathrm dA=\int_{-\pi/4}^{\pi/4}\int_{3\sqrt2\sec\theta}^6r\,\mathrm dr\,\mathrm d\theta$

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so again the total area would be $36\pi-72$ .

Or you can omit using calculus altogether and rely on some basic geometric facts. The region $R$ is a circular segment subtended by a central angle of $\dfrac\pi2$ radians. Then its area is

$\dfrac{6^2\left(\frac\pi2-\sin\frac\pi2\right)}2=9\pi-18$

so the total area is, once again, $36\pi-72$ .

An even simpler way is to subtract the area of the square from the area of the circle.

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