Well, let's imagine a peaceful little point at coordinates -3, 4. To find its distance from the x-axis, we just need to look at how far up or down it is. Since the point is 4 units above the x-axis, that's its distance. Just a happy little calculation to show how everything has its place in the world.
The distance from the y-axis is determined by the x-coordinate of a point. For the point (-1.5, 6), the distance from the y-axis is 1.5 units. Any point with an x-coordinate of either 1.5 or -1.5, such as (1.5, 4) or (-1.5, 2), will be the same distance from the y-axis as (-1.5, 6).
The line ( y = 3 ) is a horizontal line. The distance from the point ( (5, 4) ) to this line can be found by calculating the vertical distance between the point and the line. Since the y-coordinate of the point is 4 and the line is at ( y = 3 ), the distance is ( |4 - 3| = 1 ). Therefore, the distance from the point ( (5, 4) ) to the line ( y = 3 ) is 1 unit.
The unclear information given suggests that the coordinate is (-4, 0)
To find point A', which is the transformed point, you first determine the distance from point A (3, 4) to the line x = 2. The distance is the horizontal distance, which is |3 - 2| = 1 unit. Since point A' must be the same distance from the line, it can be located either at (1, 4) or (5, 4), depending on whether it is to the left or right of the line x = 2.
move to the left or right of the y- axis the given distance. place points there. move up any distance. then move to the left or right of the y- axis the given distance. place points there. now you have 4 points
The distance from the y-axis is determined by the x-coordinate of a point. For the point (-1.5, 6), the distance from the y-axis is 1.5 units. Any point with an x-coordinate of either 1.5 or -1.5, such as (1.5, 4) or (-1.5, 2), will be the same distance from the y-axis as (-1.5, 6).
Equation of circle: x^2 +y^2 -2x -6y +5 = 0 Completing the squares: (x-1)^2 +(y-3)^2 = 5 Center of circle: (1, 3) Tangent line from (3, 4) meets the x axis at: (5, 0) Distance from (5, 0) to (1, 3) = 5 using the distance formula
The line ( y = 3 ) is a horizontal line. The distance from the point ( (5, 4) ) to this line can be found by calculating the vertical distance between the point and the line. Since the y-coordinate of the point is 4 and the line is at ( y = 3 ), the distance is ( |4 - 3| = 1 ). Therefore, the distance from the point ( (5, 4) ) to the line ( y = 3 ) is 1 unit.
The unclear information given suggests that the coordinate is (-4, 0)
To find the coordinates of points on the x-axis that are 5 units away from the point (6, -3), we can use the distance formula. The distance formula is: Distance = √((x2 - x1)^2 + (y2 - y1)^2) In this case, we know that x1 = 6, y1 = -3, and distance = 5. We also know that the points are on the x-axis, so the y-coordinate is 0. So we can plug these values into the distance formula and solve for x2: 5 = √((x2 - 6)^2 + (0 - (-3))^2) 5 = √(x2 - 6)^2 + 9 25 = (x2 - 6)^2 x2 = √25 + 6 = √16 + 6 = 4 + 6 = 10 Therefore, the coordinates of the point on the x-axis that is 5 units away from (6, -3) in the positive direction of x-axis are (10, 0) and the point on the x-axis that is 5 units away from (6, -3) in the negative direction of x-axis is (2,0).
Moment of inertia about x-axis for an ellipse is = pi*b^3*a /4. Where b is the distance from the center of the ellipse to the outside tip of the minor axis. a is the distance from the ceneter of the ellipse to the outside tip of the major axis. Moment of inertia about x-axis for an ellipse is = pi*b^3*a /4. Where b is the distance from the center of the ellipse to the outside tip of the minor axis. a is the distance from the ceneter of the ellipse to the outside tip of the major axis.
move to the left or right of the y- axis the given distance. place points there. move up any distance. then move to the left or right of the y- axis the given distance. place points there. now you have 4 points
Circle equation: x^2 +y^2 -2x -6y +5 = 0 Completing the squares: (x-1)^2 +(y-3)^2 = 5 Centre of circle: (1, 3) Tangent line meets the x-axis at: (0, 5) Distance from (0, 5) to (1, 3) = 5 units using the distance formula
The answer depends on the coordinates of the point that you wish to get to. If it is (3, -4), then NO, but if it is (3, 4) then YES.
Let us consider a point (3,4). Now, the number written first is the x-coordinate and the next number is the y-coordinate. The x-axis of a graph is the horizontal line and the y-axis is the vertical line. the x-coordinates are plotted taking the x-axis for reference and the y-coordinates are plotted taking the y-axis as reference. So now, if we have the point (3,4) then we see the x-axis to see where the x-coordinate is. We have 3 as x-coordinate. Just keep a mental note of where the "3" is on the x-axis. then see the y-xis and try to find out where the y-coordinate is (In this case, 4). daw a horizontal line from 4 in the y-axis and a vertical line from 3 in the x-axis. These two lines meet a point. this point is the point we had to plot i.e., (3,4). If you observe the graph so plotted, you will see that (3,4) is at a distance of 3 units from the origin along the x-axis and 4 units from the origin along the y-axis.
1
Zero, the line is parallel to the x-axis at a distance 4+3 = 7