(Distance)2 = (2 - 5)2 + (6 - 2)2
To find the distance between the points (-2, 5) and (-2, 0), we can use the distance formula. Since both points have the same x-coordinate (-2), the distance is simply the difference in their y-coordinates: |5 - 0| = 5. Therefore, the distance between the two points is 5 units.
Points: (1, -2) and (1, -5) Distance: 3 units by using the distance formula
To find the distance between the points (7, 5) and (4, 9), you can use the distance formula: ( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ). Plugging in the values, ( d = \sqrt{(4 - 7)^2 + (9 - 5)^2} = \sqrt{(-3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 ). Therefore, the distance between the two points is 5 units.
Distance from (0, 0) to (5, 12) using distance formula is 13
The distance between two points in a plane can be found using the distance formula: sqrt((x2 - x1)^2 + (y2 - y1)^2). In this case, the distance between the points (-1, 2) and (2, 6) is sqrt((2 - (-1))^2 + (6 - 2)^2) = sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5.
To find the distance between the points (-2, 5) and (-2, 0), we can use the distance formula. Since both points have the same x-coordinate (-2), the distance is simply the difference in their y-coordinates: |5 - 0| = 5. Therefore, the distance between the two points is 5 units.
Points: (1, -2) and (1, -5) Distance: 3 units by using the distance formula
To find the distance between two points in 3D space (x1, y1, z1) and (x2, y2, z2), use the distance formula: Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2) Substitute the coordinates into the formula to find the distance.
To find the distance between the points (7, 5) and (4, 9), you can use the distance formula: ( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ). Plugging in the values, ( d = \sqrt{(4 - 7)^2 + (9 - 5)^2} = \sqrt{(-3)^2 + (4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 ). Therefore, the distance between the two points is 5 units.
Distance from (0, 0) to (5, 12) using distance formula is 13
If you mean points of (-2, 4) and (5, 4) then using the distance formula it is 7
Apply Pythagoras d^(2) = (x^(2) + y^(2) Hence for the points (-3,2)& ( 5. -1) d^(2) = (-3-5)^(2) + (2 - - 1)^(2) (Note the double minus). d^(2) = (-8)^(2) + (3)^(2) d^(2) = 64 + 9 d^(2) = 73 d = sqrt(73) d = 8.54400....
To find the distance between the points (1, 4) and (-5, 7), you can use the distance formula: ( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ). Plugging in the coordinates, we get ( d = \sqrt{((-5) - 1)^2 + (7 - 4)^2} = \sqrt{(-6)^2 + (3)^2} = \sqrt{36 + 9} = \sqrt{45} ). Therefore, the distance between the points is ( 3\sqrt{5} ) or approximately 6.71 units.
The distance between two points in a plane can be found using the distance formula: sqrt((x2 - x1)^2 + (y2 - y1)^2). In this case, the distance between the points (-1, 2) and (2, 6) is sqrt((2 - (-1))^2 + (6 - 2)^2) = sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5.
To find the distance between the points (2, 5) and (-4, 8), you can use the distance formula: [ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ] Plugging in the coordinates, the expression becomes: [ d = \sqrt{((-4) - 2)^2 + (8 - 5)^2} = \sqrt{(-6)^2 + (3)^2} = \sqrt{36 + 9} = \sqrt{45} ] Thus, the distance is ( \sqrt{45} ) or ( 3\sqrt{5} ).
To find the distance between the points (7, 6) and (-5, 3), you can use the distance formula: (d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}). Plugging in the values, we get (d = \sqrt{(-5 - 7)^2 + (3 - 6)^2} = \sqrt{(-12)^2 + (-3)^2} = \sqrt{144 + 9} = \sqrt{153}). The distance is approximately 12.25 units.
Points: (6, 5) and (30, 15) Distance: 26 by using the distance formula