SOLUTION: I would mean alot to me if someone could help me out on this problem I did find a smiliar question to mine, but could really understand how i would get the answer. Consider t

Algebra ->  Parallelograms -> SOLUTION: I would mean alot to me if someone could help me out on this problem I did find a smiliar question to mine, but could really understand how i would get the answer. Consider t      Log On


   



Question 158027: I would mean alot to me if someone could help me out on this problem
I did find a smiliar question to mine, but could really understand how i would get the answer.
Consider the points P(10,2),Q(1,11), and R(-8,2) on a coordinate plane. Where must the point S be located so that the quadrilateral PQRS is a square?
i found the slope of PQ i got -1
I found the slope of QR i got 1
then what do we do after gettting the slope
please help

Answer by Electrified_Levi(103) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, Hope I can help,
.
I would mean alot to me if someone could help me out on this problem
I did find a smiliar question to mine, but could really understand how i would get the answer.
Consider the points P(10,2),Q(1,11), and R(-8,2) on a coordinate plane. Where must the point S be located so that the quadrilateral PQRS is a square?
i found the slope of PQ i got -1
I found the slope of QR i got 1
then what do we do after gettting the slope
please help
.
First we need to plot the three points on a graph
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Here are the 3 points
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(P)(10,2) = red
(Q)(1,11) = green
(R)(-8,2) = blue
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We will need to find two equations,
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1.equation of the line from the unknown point(x,y) to (-8,2)
2.equation of the line from the unknown point(x,y) to (10,2)
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The line from the unknown point (x,y) to (-8,2) is parallel to the line from (10,2) to (1,11)(Line PQ)(redish brown line)(line has slope of (-1))
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Since we are trying to find the equation of the line (contains unknown point(x,y) and (-8,2)), the line we are trying to find is parallel to the line with the slope of (-1), since parallel lines have the same slope, we know our unknown line has the slope of (-1), we can replace "m" with (-1) in our slope intercept equation, +y+=+mx+%2B+b+, where "m" is the slope, "b" is the y intercept
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+y+=+mx+%2B+b+ = +y+=+%28-1%29x+%2B+b+ = +y+=+-x+%2B+b+, since we have a point on the line (-8,2), we can replace "x" and "y" with (-8,2)(x,y)
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+y+=+-x+%2B+b+ = +%282%29+=+-%28-8%29+%2B+b+ = +2+=+8+%2B+b+
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To solve for "b" we will move "8" to the left side
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+2+=+8+%2B+b+ = +2+-+8+=+8+-+8+%2B+b+ = +2+-+8+=+b+ = +%28-6%29+=+b+ = +b+=+%28-6%29+
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"b" = (-6), we can replace "b" with (-6) in our equation
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+y+=+-x+%2B+b+ = +y+=+-x+%2B+%28-6%29+ = +y+=+-x+-+6+
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We can check by replacing "x" and "y" with (-8,2)(x,y)
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(-8,2)+y+=+-x+-+6+ = +2+=+-%28-8%29+-+6+ = +2+=+8+-+6+ = +2+=+2+ (True)
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One of our two equation we have to solve is +y+=+-x+-+6+(Line RS),
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we will now solve for the second unknown line equation
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2. equation of the line that contains the unknown point(x,y) and (10,2)
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This line is parallel to the line that contains (-8,2) and (1,11)(Line QR)(green line)(line that has slope of (1)).
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The line that contains the unknown point and (10,2) is parallel to the line +y+=+x+%2B+10+, (They both have the same slope)
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The line that contains the unknown point and (10,2) has a slope of "1", lets replace "m" with "1" in our slope intercept equation
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+y+=+mx+%2B+b+ = +y+=+%281%29x+%2B+b+ = +y+=+x+%2B+b+
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Since the line contains (10,2) we can replace "x" and "y" with (10,2)(x,y)
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+y+=+x+%2B+b+ = +%282%29+=+%2810%29+%2B+b+ = +2+=+10+%2B+b+, we will move "10" to the left side to solve "b"
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+2+=+10+%2B+b+ = +2-10+=+10+-+10+%2B+b+ = +2-10+=+b+ = +%28-8%29+=++b+ = +b+=+%28-8%29+, we can replace "b" with (-8) in our equation
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+y+=+x+%2B+%28-8%29+ = +y+=+x+-+8+, We can check by replacing "x" and "y" with (10,2)
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(10,2), +y+=+x+-+8+ = +%282%29+=+%2810%29+-+8+ = +2+=+10+-+8+ = +2+=+2+ (True)
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(Line QS)+y+=+x+-+8+ is our second equation we had to solve
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We found our two unknown equations(our two unknown line equations)
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+y+=+x+-+8+ (Line QS)
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+y+=+-x+-+6+ (Line RS)
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We can now solve for both "x" and "y" to get our unknown point, this is the way I solve systems of equations
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First solve for a letter(doesn't matter which one)(since we don't have to do anything to solve for "y"(it is solved already for us), we will put the two answers that "y" is together into an equation(since "y" is one number, both of the answers equal each other)
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Here are our two answers
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++x+-+8+
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+-x+-+6+
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Lets put them together
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++%28x+-+8%29+=+%28-x+-+6%29+ = ++x+-+8+=+%28-x%29+-+6+, we will move (-x) to the left side
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++x+-+8+=+%28-x%29+-+6+ = ++x+%2B+x+-+8+=+%28-x%29+%2B+x+-+6+ = ++2x+-+8+=%28-+6%29+, We will move (-8) to the right side to solve "x" even more
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++2x+-+8+=%28-+6%29+ = ++2x+-+8+%2B+8+=%28-+6%29+%2B+8+ = ++2x+=+2+
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We will divide each side by "2" to get "x"
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++2x+=+2+ = ++2x%2F2+=+2%2F2+ = ++x+=+2%2F2+ ++x+=+1+
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"x" = 1, we can replace "x" with "1" in one of the two equations to get "y"
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+y+=+x+-+8+
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+y+=+-x+-+6+
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We will use the first equation
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+y+=+x+-+8+ = +y+=+%281%29+-+8+ = +y+=+1+-+8+ = +y+=+%28-7%29+
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"y" = (-7), we can check our answers by replacing "x" and "y" in our two equations
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"x" = "1"
"y" = (-7)
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+y+=+x+-+8+ = +%28-7%29+=+%281%29+-+8+ = +%28-7%29+=+1+-+8+ = +%28-7%29+=+%28-7%29+ (True)
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+y+=+-x+-+6+ = +%28-7%29+=+-%281%29+-+6+ = +%28-7%29+=+-1+-+6+ = +%28-7%29+=+%28-7%29 ( True)
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"x" = "1"
"y" = (-7)
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Since points are given as (x,y), our unknown point(Point "S") is (1,-7)
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You can find this answer by a shortcut
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Since this is a square, the diagonals are the same, you could measure the distance between (-8,2) and (10,2)(18 squares), you would just go from (1,11) and move down 18 squares, and you would find the other point
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Unknown point( Point "S") = (1,-7)
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Hope I helped, Levi