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M is the midpoint of pq verify that this is the midpoint by using the distance formula to show tha pm equals mq?
Let P(x1, y1), Q(x2, y2), and M(x3, y3).If M is the midpoint of PQ, then,(x3, y3) = [(x1 + x2)/2, (y1 + y2)/2]We need to verify that,√[[(x1 + x2)/2 - x1]^2 + [(y1 + y2)/2 - y1]^2] = √[[x2 - (x1 + x2)/2]^2 + [y2 - (y1 + y2)/2]^2]]Let's work separately in both sides. Left side:√[[(x1 + x2)/2 - x1]^2 + [(y1 + y2)/2 - y1]^2]= √[[(x1/2 + x2/2)]^2 - (2)(x1)[(x1/2 + x2/2)) + x1^2] + [(y1/2 + y2/2)]^2 - (2)(y1)[(y1/2 + y2/2)] + y1^2]]= √[[(x1)^2]/4 + [(x1)(x2)]/2 + [(x2)^2]/4 - (x1)^2 - (x1)(x2) + (x1)^2 +[(y1)^2]/4 + [(y1)(y2)]/2 + [(y2)^2]/4 - (y1)^2 - (y1)(y2) + (y1)^2]]= √[[(x1)^2]/4 - [(x1)(x2)]/2 + [(x2)^2]/4 + [(y1)^2]/4 - [(y1)(y2)]/2 + [(y2)^2]/4]]Right side:√[[x2 - (x1 + x2)/2]^2 + [y2 - (y1 + y2)/2]^2]]= √[[(x2)^2 - (2)(x2)[(x1/2 + x2/2)] + [(x1/2 + x2/2)]^2 + [(y2)^2 - (2)(y2)[(y1/2 + y2/2)] + [(y1/2 + y2/2)]^2]]= √[[(x2)^2 - (x1)(x2) - (x2)^2 + [(x1)^2]/4 + [(x1)(x2)]/2 + [(x2)^2]/4 + (y2)^2 - (y1)[(y2) - (y2)^2 + [(y1)^2]/4) + [(y1)(y2)]/2 + [(y2)^2]/4]]= √[[(x1)^2]/4 - [(x1)(x2)]/2 + [(x2)^2]/4 + [(y1)^2]/4 - [(y1)(y2)]/2 + [(y2)^2]/4]]Since the left and right sides are equals, the identity is true. Thus, the length of PM equals the length of MQ. As the result, M is the midpoint of PQ
How do you solve for slope from two points and a graph?
Points: (x1, y1) and (x2, y2) Slope: y1-y2/x1-x2
What is the mathematical equation of slope?
The slope between two points, (x1, y1) and (x2, y2) is: (y1 - y2) / (x1 - x2)
How do you find linear equations with just coordinates?
if we take the (x1,y1),(x2,y2) as coordinates the formula was (x-x1)/(x2-x1)=(y-y1)/(y2-y1)
How do you get the distance?
The distance between two points, (x1,y1) , (x2,y2) = squareroot[(x2-x1)2 + (y2-y1)2]
How do you draw a square using line command?
Line (x1, y1, x2, y1); Line (x2, y1, x2, y2); Line (x2, y2, x1, y2); Line (x1, y2, x1, y1);
When using the slope formula does it have to be (y2-y1)(x2-x1) or can it be (y1-y2)(x1-x2)?
Oh, don't you worry, friend! When using the slope formula, whether you do (y2-y1)/(x2-x1) or (y1-y2)/(x1-x2), the answer will be the same! It's all about the difference in vertical and horizontal values, and as long as you stay consistent in your calculations, you'll find the slope just fine. Just trust your instincts and enjoy the process of solving the equation.
X1 plus x2 divided by 2 y1 plus y2 divided by 2?
Oh honey, you've got yourself a classic case of finding the average of two points in a coordinate plane. All you need to do is add the x-coordinates (x1 + x2) and divide by 2 to get the x-coordinate of the midpoint. Then do the same for the y-coordinates (y1 + y2), divide by 2, and voila, you've got the y-coordinate of the midpoint. Easy peasy lemon squeezy!
M is the midpoint of pq verify that this is the midpoint by using the distance formula to show tha pm equals mq?
Let P(x1, y1), Q(x2, y2), and M(x3, y3).If M is the midpoint of PQ, then,(x3, y3) = [(x1 + x2)/2, (y1 + y2)/2]We need to verify that,√[[(x1 + x2)/2 - x1]^2 + [(y1 + y2)/2 - y1]^2] = √[[x2 - (x1 + x2)/2]^2 + [y2 - (y1 + y2)/2]^2]]Let's work separately in both sides. Left side:√[[(x1 + x2)/2 - x1]^2 + [(y1 + y2)/2 - y1]^2]= √[[(x1/2 + x2/2)]^2 - (2)(x1)[(x1/2 + x2/2)) + x1^2] + [(y1/2 + y2/2)]^2 - (2)(y1)[(y1/2 + y2/2)] + y1^2]]= √[[(x1)^2]/4 + [(x1)(x2)]/2 + [(x2)^2]/4 - (x1)^2 - (x1)(x2) + (x1)^2 +[(y1)^2]/4 + [(y1)(y2)]/2 + [(y2)^2]/4 - (y1)^2 - (y1)(y2) + (y1)^2]]= √[[(x1)^2]/4 - [(x1)(x2)]/2 + [(x2)^2]/4 + [(y1)^2]/4 - [(y1)(y2)]/2 + [(y2)^2]/4]]Right side:√[[x2 - (x1 + x2)/2]^2 + [y2 - (y1 + y2)/2]^2]]= √[[(x2)^2 - (2)(x2)[(x1/2 + x2/2)] + [(x1/2 + x2/2)]^2 + [(y2)^2 - (2)(y2)[(y1/2 + y2/2)] + [(y1/2 + y2/2)]^2]]= √[[(x2)^2 - (x1)(x2) - (x2)^2 + [(x1)^2]/4 + [(x1)(x2)]/2 + [(x2)^2]/4 + (y2)^2 - (y1)[(y2) - (y2)^2 + [(y1)^2]/4) + [(y1)(y2)]/2 + [(y2)^2]/4]]= √[[(x1)^2]/4 - [(x1)(x2)]/2 + [(x2)^2]/4 + [(y1)^2]/4 - [(y1)(y2)]/2 + [(y2)^2]/4]]Since the left and right sides are equals, the identity is true. Thus, the length of PM equals the length of MQ. As the result, M is the midpoint of PQ
What is y2-y1x2-x1?
The equation (y2-y1)/(x2-x1) is known as the point-slope formula. It gives the slope for a line given two points of coordinatesÊ(x1, y1) and (x2, y2).
What is the formula to calculate the slope?
It's m = y2 - y1/ x2- x1 It's m equals y2 minus y1 over x2 minus x1
How do you solve for slope from two points and a graph?
Points: (x1, y1) and (x2, y2) Slope: y1-y2/x1-x2
What is the mathematical equation of slope?
The slope between two points, (x1, y1) and (x2, y2) is: (y1 - y2) / (x1 - x2)
How do you find linear equations with just coordinates?
if we take the (x1,y1),(x2,y2) as coordinates the formula was (x-x1)/(x2-x1)=(y-y1)/(y2-y1)
What formula is used to find the length between two points in a coordinate plane?
It is the Pythagorean distance formmula.If P = (x1, y1) and Q = (x2, y2) thenDistance between P and Q = sqrt[(x1 - x2)2 + (y1 - y2)2]It is the Pythagorean distance formmula.If P = (x1, y1) and Q = (x2, y2) thenDistance between P and Q = sqrt[(x1 - x2)2 + (y1 - y2)2]It is the Pythagorean distance formmula.If P = (x1, y1) and Q = (x2, y2) thenDistance between P and Q = sqrt[(x1 - x2)2 + (y1 - y2)2]It is the Pythagorean distance formmula.If P = (x1, y1) and Q = (x2, y2) thenDistance between P and Q = sqrt[(x1 - x2)2 + (y1 - y2)2]
How do you get the distance?
The distance between two points, (x1,y1) , (x2,y2) = squareroot[(x2-x1)2 + (y2-y1)2]
How do you do equations and graphs for slope?
The equation for the slope between the points A = (x1, y1) and B = (x2, y2) = (y2 - y1)/(x2 - x1), provided x1 is different from x2. If x1 and x2 are the same then the slope is not defined.
