Question

Find a cubic Hermite curve as Curve B in both the matrix and algebraic forms. vectors by a factor of 6 while keeping the directions and the starting and ending points unchanged, find the modified curve as Curve C in both the matrix and algebraic forms.

Answer 1

To find all points on the x-axis that are units away from the point (0, 0), we can use the point-slope form of a linear equation. The point-slope form of a linear equation is given by y - y1 = m(x - x1), where (x1, y1) is a reference point and m is the slope of the line

Explanation:

To find all points on the x-axis that are units away from the point (0, 0), we can use the point-slope form of a linear equation. The point-slope form of a linear equation is given by y - y1 = m(x - x1), where (x1, y1) is a reference point and m is the slope of the line.

In this case, the point (0, 0) is our reference point, and the distance away from the x-axis is 1 unit. Therefore, we can use the point-slope form of the equation for the x-axis, which is y - y1 = 0(x - 0).

Now, we can find all points on the x-axis that are units away from the point (0, 0). To do this, we can substitute different x-coordinates into the equation and solve for the corresponding y-coordinates. For example, if x = 1, we get y = 1, and the point is (1, 1). If x = -1, we get y = -1, and the point is (-1, -1). If x = 2, we get y = 2, and the point is (2, 2). If x = -2, we get y = -2, and the point is (-2, -2).

In summary, all points on the x-axis that are units away from the point (0, 0) are (1, 1), (-1, -1), (2, 2), and (-2, -2).