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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
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 solve for slope from two points and a graph?
Points: (x1, y1) and (x2, y2) Slope: 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);
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
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 solve for slope from two points and a graph?
Points: (x1, y1) and (x2, y2) Slope: 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.
How do you find the vector sum and vector diffirences of two vector quantities?
If a vector is given in component form <x1,y1> and <x2,y2>, then you add or subtract the corresponding components. <x1,y1>+<x2,y2>=<x1+x2,y1+y2>
Can you evaluate 4xy for x1 y2?
It is 8.
