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The distance is 2, along the x-value line x=3 i.e. (3, y).
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Distance between point A 5 7 and point B 2 4?
|AB| = sqrt[(5 - 2)2 + (7 - 4)2] =sqrt[9 + 9] = 3*sqrt(2)
What is the transformation of B(2 4) when dilated with a scale factor of ½ using the point (4 6) as the center of dilation?
When doing enlargements through a centre, the new position of any point is the distance of that point from the centre multiplied by the scale factor; it is easiest to treat the x- and y- coordinates separately.To enlarge (2, 4) by a scale factor of ½ with (4, 6) as the centre of enlargement:x: distance is (4 - 2) = 2 → new distance is 2 × ½ = 1 → new x is 2 + 1 = 3y: distance is (6 - 4) = 2 → new distance is 2 × ½ = 1 → new y is 4 + 1 = 5→ (2, 4) when enlarged by a scale factor of ½ with a centre of (4, 6) transforms to (3, 5).
What is the distance from the centre of a circle to a point on the x axis that forms a tangent line when it meets the circle of x2 plus y2 -2x -6y plus 5 equals 0 at the point 3 and 4?
Point of contact: (3, 4) Circle equation: x^2 +y^2 -2x -6y+5 = 0 Completing the squares: (x-1)^2 +(y-3)^2 -1 -9 +5 = 0 So: (x-1)^2 +(y-3) = 5 Centre of circle: (1, 3) Slope of radius: (3-4)/(1-3) = 1/2 Slope of tangent: -2 Equation of tangent line: y-4 = -2(x-3) => 2x+y = 10 Tangent line meets the x axis at: (5, 0) Using formula distance from (1, 3) to (5, 0) = 5 units
What is the distance between the points -1 -9 and 4 -2?
Points: (-1, -9) and (4, -2) Distance: (-1-4)2+(-9--2)2 = 74 and the square root of this is the distance which is about 8.602 to 3 decimal places
What is the perpendicular distance from the point 2 and 4 to the straight line equation of y equals 2x plus 10 showing key stages of work?
1 Coordinates: (2, 4) 2 Equation: y = 2x+10 3 Perpendicular equation: y = -0.5+5 4 They intersect at: (-2, 6) 5 Distance is the square root of: (-2, -2)2+(6, -4) = 2*sq rt of 5 = 4.472 to 3 decimal places
What is the distance between point negative 6 1 and point 4 3?
It is the square root of (-6-4)2+(1-3)2 = 2 times sq rt of 26 or about 10.198 to 3 decimal places
on a coordinate plane, what is the distance between the point (-2, -3) and (-2, 10)?
If you mean points of (5, 5) and (1, 5) then the distance is 4
Distance between point A 5 7 and point B 2 4?
|AB| = sqrt[(5 - 2)2 + (7 - 4)2] =sqrt[9 + 9] = 3*sqrt(2)
What is the transformation of B(2 4) when dilated with a scale factor of ½ using the point (4 6) as the center of dilation?
When doing enlargements through a centre, the new position of any point is the distance of that point from the centre multiplied by the scale factor; it is easiest to treat the x- and y- coordinates separately.To enlarge (2, 4) by a scale factor of ½ with (4, 6) as the centre of enlargement:x: distance is (4 - 2) = 2 → new distance is 2 × ½ = 1 → new x is 2 + 1 = 3y: distance is (6 - 4) = 2 → new distance is 2 × ½ = 1 → new y is 4 + 1 = 5→ (2, 4) when enlarged by a scale factor of ½ with a centre of (4, 6) transforms to (3, 5).
What is the distance between the points -4 3 and 3 -1?
Points: (-4, 3) and (3, -1) Distance: (3--4)2+(-1-3)2 = 65 and the square root if this is the distance which is just over 8
The distance from one point to another is?
The distance formula is Example: (1,1) and (3,2) X1=1 X2=3 Y1=1 Y2=2 (3-1)2=4 and (2-1)2=1 4+1=5 The answer is going to be the square root of 5
What is the distance from point a -8 4 to point b 5 4?
Using Pythagoras: distance = √(change_in_x2 + change_in_y2) = √((5 - -8)2 + (4 - 4)2) = √(132 + 02) = √(132) = 13 units.
What is the distance between point -3 5 and point 4 -6 in the coordinate plane?
The answer depends on the metric used. The Euclidean distance is sqrt[(-3-4)2 + (5+6)2] = sqrt[72 + 112] =sqrt(49 + 121) = sqrt(170) = 13.0384 (to 6 sf). The Minkowsky distance, on the other hand, is |-3-4| + |5+6| = 7 + 11 = 18. There are other metrics.
What is the distance from point A -6 4 to point B 3 -8?
15 = sq rt of (81+144)Distance formula between two points (in the plane): d = √[(x2 - x1)^2 + (y2 - y1)^2]Let A be the point with cordinates (x1, y1), and B the point with coordinates (x2, y2).Substitute the given values into that formula:d = √[(x2 - x1)^2 + (y2 - y1)^2]d = √[(3 - -6)^2 + (-8 - 4)^2]d = √[9^2 + (-12)^2]d = √[81 +144)d = √225d = 15Thus, the distance from point A to point B is 15.
What is the distance of point -3 4 from x axis?
5 units
How can you find the distance between two numbers on a number line?
Just subtract the lowest number from the greatest number. For example, the distance between 3 and 8, is 8 - 3 = 5 units, the distance between -2 and 3, is 3 - (-2) = 3 + 2 = 5 units, the distance between -4 and -2, is -2 - (-4) = -2 + 4 = 2 units.
Find the distance between the following pairs of points Round your answer to the nearest tenth 4 and -3 and the second is 2 and 4?
P1 = (4, -3)P2 = (2, 4)(Distance)2 = (delta-x)2 + (delta-y)2(Distance)2 = (4 - 2)2 + (-3 - 4)2 = (2)2 + (-7)2 = 4 + 49 = 53Distance = sqrt(53) = 7.3 (rounded)
