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N76 [4]
3 years ago
6

Which statement proves that the diagonals of square PQRS are perpendicular bisectors of each other? The length of SP, PQ, RQ, an

d SR are each 5. The slope of SP and RQ is and the slope of SR and PQ is . The length of SQ and RP are both . The midpoint of both diagonals is , the slope of RP is 7, and the slope of SQ is .
Mathematics
2 answers:
MatroZZZ [7]3 years ago
8 0

Answer: the correct answer is letter D, I just took the test and got it right on edg

pochemuha3 years ago
6 0

Answer:

Step-by-step explanation:

Given is a square PQRS.  Its diagonals are PR and QS.

To prove that PR is perpendicular to QS

We have been given as

SP = PQ =RQ =SR =5.

Using Pythagorean theorem we have

both diagonals will equal

5\sqrt2

If P is the origin, (say) then coordinates of vertices would be (0,0) (5,0) (5,5) and (0,5)

Mid point would be (2.5,2.5) for both

Slope of RP = change in y coordinate/change in x coordinate

= 1

Slope of QS = 5/-5 = -1

Product of the two slopes = -1

Hence the two diagonals are perpendicular

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What is the equation of a line that is parallel to the line 2x + 5y = 10 and passes through the point (–5, 1)? Check all that ap
Bond [772]

So the right options are:

y=-\frac{2}{5}x-1

2x+5y = -5

y-1=-\frac{2}{5}(x+5)

Further explanation:

Given equation of line is:

2x+5y=10

We have to convert it into point-slope form

2x+5y=10\\5y=-2x+10\\Dividing\ both\ sides\ by\ 5\\y=-\frac{2}{5}x+\frac{10}{5}\\y=-\frac{2}{5}x+2

The co-efficient of x is the slope of the line

So,

m= -\frac{2}{5}

As the required line is parallel to given line, it will also have same slope.

Let m1 be the slope of required line

Then the line will be:

y=m_1x+b

Putting the value of slope

y=\frac{2}{5}x+b

Putting (-5,1) in the equation to find the value of b

1=-\frac{2}{5}(-5)+b\\1=2+b\\b=1-2\\b=-1

Putting the values of slope and b in equation

y=-\frac{2}{5}x-1

Multiplying the whole equation by 5 will give us:

5y = -2x-5

2x+5y = -5

Another form of equation of line is Point-slope form

y-y_1=m(x-x_1)

Putting the values of slope and point in the equation, we get

y-1=-\frac{2}{5}(x+5)

So the right options are:

y=-\frac{2}{5}x-1

2x+5y = -5

y-1=-\frac{2}{5}(x+5)

Keywords: Point-Slope form, Parallel lines

Learn more about point slope form at:

  • brainly.com/question/4464845
  • brainly.com/question/4522984

#LearnwithBrainly

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