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sammy [17]
2 years ago
4

A, B, C, and D have the coordinates (-8, 1), (-2,4),(-3,-1) and (-6,5), respectively. Which sentence about the points is true?

Mathematics
1 answer:
Y_Kistochka [10]2 years ago
5 0

Answer:

B. ^\leftrightarrow_{AB} and ^\leftrightarrow_{CD}  are perpendicular lines.

Step-by-step explanation:

We can quickly plot the points in the cartesian plane as shown in the attachment.

A visual representation will help us see that A,B,C, and D do not lie on the same line.

The slope of AB  is \frac{4-1}{-2--8} =\frac{3}{6}=\frac{1}{2}

The slope of CD is \frac{5--1}{-6--3} =\frac{6}{-3}=-2

The two slopes are negative reciprocals of each other.

It is true that line AB and line CD are perpendicular.

These two lines cannot be perpendicular and parallel at the same time.

It is also not possible that, the two lines are perpendicular but will not intersect

Therefore the correct choice is B

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Hey ^-^ can someone please help me with this problem:
soldi70 [24.7K]

Answer:   8 (Pi - sqrt(3))

Discussion:

The area of the shaded region is that of the semicircle minus the area of the triangle..

Area of semicircle = 1/2 * Pi * R^2        

   Where R^2 is the square of the radius of the circle. In our case, R ( = OC)

    = 4 so the semicircle area is

   (1/2) * Pi * (4^2) = (1/2) * Pi * 16 = 8 Pi

Area of triangle.

  First of all, angle ACB is a right angle ( i.e. 90 degrees).

    * This is the Theorem of Thales from elementary Plane Geometry. *

 so by Pythagoras

   AC^2 + BC^2 = AB^2

But CB = 4 (given) and AB = 4*2 = 8 ( the diameter is twice the radius).

Substituting these in Pythagoras gives

   AC^2 + 4^2 = 8^2 or

   AC^2 = 8^2 - 4^2- = 64 - 16 = 48

   Hence AC = sqrt(48) = sqrt (16*3) = 4 * sqrt(3)

We are almost done! The area of the triangle is given by

  (1/2) b * h = (1/2)  BC * AC = (1/2) 4 * (4 * sqrt(3)) =  8 sqrt(3)

We conclude the area area of the shaded part is

 8 PI - 8 sqrt(3)   = 8 (Pi - sqrt(3))

Note that sqrt(3) is approx  1.7 so (PI - sqrt(3)) is a positive number, as it better well be!

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