First let's look at what y=2x+2 looks like. It is the equation of a line with slope 2 and y intercept 2. When we reflect across the y axis, the point (x,y) become (-x,y). Remember that when you reflect across the x axis (x,y) become (x,-y) and when you reflect across the y axis (x,y) becomes (-x,y) So one way to do this problem is algebraically. The slope cannot change values but must change sign with the reflection about the y axis. Let's take a point, not on the y axis, but on the line. (the point (0,2) is the one on the y axis and it would not change). Take (1,4) and note that after reflection the point is (-1,4). So we have a line with point (-1,4) and slope -2. Put this in point slope form and we have y-4=-2(x+1) or y-4=-2x-2 which in slope intercept form is y=-2x+2. The equation of the reflected line is y=-2x+2 So the original line was y=2x+2 and the reflected line is y=-2x+2. The y intercept is the same. Look at a few points. The point (5,12) was on the original line, and (-5,12) is on the reflected line. (2,4) was on the original one and (-2,4) is on the new one. We could have done this all geometrically without the algebra. The line we started with was y=2x+2. If we reflect across the y axis, we cannot change the y intercept, but the slope must become it's negative. The y values remain unchanged, Right away we could write the equation y=-2x+2. This would change the slope to -2 and the y intercept would remain the same, 2. You can also graph the original line on graph paper, reflect it and note the new slope and the y intercept. This would be another way to find the answer. Sometimes we want the value of points when reflected across axes or even arbitrary lines. Matrices are often used for this. The matrix below will transform a point across the y axis. |-1 0x| |0 1y|
Measure the distance between the point where the line intercepts the Y axis and the origin
Measure the angle in a plane perpendicular to the axis of rotation, between the position of a fixed point at the start and end of the rotation.
Having one given point is not enough. A y-intercept is described as the point of intersection of a function or relation or line and the ordinate axis (or y-axis).Suppose a function intersects the y-axis at (0, 6), then 6 is the value of the y-intercept. Or if the line that passes through (5, 4) is parallel to x-axis, then y-intercept is 4; if it passes through the origin, then y-intercept is 0; if it is perpendicular to x-axis (or parallel to y-axis) there is not an y-intercept.
2/3,0.5/3
If the coordinates of a point, before reflection, were (p, q) then after reflection, they will be (-p, q).
The answer is b.
no
The answer is the x coordinate of the point.
y/x where y is the distance of point from x axis and x is the distance from y axis
Draw a line joining a point and its image and find its midpoint. Repeat for another pair of point and its image. The line joining these midpoints is the line of reflection.
First let's look at what y=2x+2 looks like. It is the equation of a line with slope 2 and y intercept 2. When we reflect across the y axis, the point (x,y) become (-x,y). Remember that when you reflect across the x axis (x,y) become (x,-y) and when you reflect across the y axis (x,y) becomes (-x,y) So one way to do this problem is algebraically. The slope cannot change values but must change sign with the reflection about the y axis. Let's take a point, not on the y axis, but on the line. (the point (0,2) is the one on the y axis and it would not change). Take (1,4) and note that after reflection the point is (-1,4). So we have a line with point (-1,4) and slope -2. Put this in point slope form and we have y-4=-2(x+1) or y-4=-2x-2 which in slope intercept form is y=-2x+2. The equation of the reflected line is y=-2x+2 So the original line was y=2x+2 and the reflected line is y=-2x+2. The y intercept is the same. Look at a few points. The point (5,12) was on the original line, and (-5,12) is on the reflected line. (2,4) was on the original one and (-2,4) is on the new one. We could have done this all geometrically without the algebra. The line we started with was y=2x+2. If we reflect across the y axis, we cannot change the y intercept, but the slope must become it's negative. The y values remain unchanged, Right away we could write the equation y=-2x+2. This would change the slope to -2 and the y intercept would remain the same, 2. You can also graph the original line on graph paper, reflect it and note the new slope and the y intercept. This would be another way to find the answer. Sometimes we want the value of points when reflected across axes or even arbitrary lines. Matrices are often used for this. The matrix below will transform a point across the y axis. |-1 0x| |0 1y|
To find the x-coordinate of a point on the xy-plane, you look at the horizontal distance of the point from the y-axis. The y-coordinate of a point on the xy-plane is the vertical distance of the point from the x-axis.
What is the value of x, to the nearest degree? 1 point Captionless Image
Your reflection is based on the shape the mirror is.Example:If the mirror was bent outwards the reflected image would appear wider.If the mirror was bent inwards the reflected image would appear thinner.Usually you can commonly find mirrors like these in fun houses, and can be found in various shapes such as cones etc.
The x-intercept is the point where the line crosses the x-axis. 'y' is zero at every point on the x-axis. So to find where the line crosses the x-axis, make 'y' zero, and solve the equation for 'x'. -- The y-intercept is the point where the line crosses the y-axis. 'x' is zero at every point on the y-axis. So to find where the line crosses the y-axis, make 'x' zero, and solve the equation for 'y'. This is the central idea that the following answer calls "zero out". =========================================================X and Y intercepts
-find the area(A) of the shape above the neutral axis (or above a particular point if given) - locate the centroid (y')of the shape relative to the neutral axis(or above point) using y' = ∑AiYi / ∑Ai - first moment of area = A*y' (or y' + distance of given point from neutral axis)