State the normal of a line that is parallel to 2x-4y+5
2x-4y+5=a so rearranging,
-4y=a-2x-5 or
y=-a/4+2x/4+5/4
The slope of this line is 2/4. A line normal to this would have a slope of minus the inverse, or -4/2, or -2. So the equation of normal line would be:
y=-2x
To determine the normal of a line parallel to a given line, we need to find the slope of the given line. The slope-intercept form of a line is y = mx + b, where m is the slope and b is the y-intercept.
Given the equation of the line 2x - 4y + 5 = 0, we can rearrange it to solve for y:
2x - 4y + 5 = 0
-4y = -2x - 5
y = (2/4)x + (5/(-4))
y = (1/2)x - (5/4)
Now we can identify the slope of the given line, which is (1/2). Since we want a line parallel to this, the slope of the new line will also be (1/2).
The slope of a line perpendicular to another line is the negative reciprocal of the slope of the given line. So, the slope of the normal line will be the negative reciprocal of (1/2).
To find the negative reciprocal, we invert the fraction and change its sign. Thus, the negative reciprocal of (1/2) is -2.
Therefore, the normal of a line parallel to 2x - 4y + 5 is -2.