what is line perpendicular to line 5x+y=57 that goes through the point (10,6).
To find the line perpendicular to the given line and passing through the point (10,6), we can follow these steps:
Step 1: Write the equation of the given line in slope-intercept form (y = mx + b).
5x + y = 57
Rearrange the equation to solve for y:
y = -5x + 57
Now we have the equation of the given line in slope-intercept form, where the coefficient of x (-5) is the slope of the line.
Step 2: Determine the slope of the line perpendicular to the given line.
Perpendicular lines have slopes that are negative reciprocals of each other. Therefore, the slope of the perpendicular line would be the negative reciprocal of -5.
The negative reciprocal of -5 is 1/5, which means that the perpendicular line will have a slope of 1/5.
Step 3: Use the point-slope form to find the equation of the perpendicular line.
We have the slope (m = 1/5) and a point (10,6) through which the line passes. We can now use the point-slope form of a line:
y - y1 = m(x - x1)
Substitute the values of (x1, y1) = (10, 6) and m = 1/5 into the equation:
y - 6 = 1/5(x - 10)
Now simplify the equation:
y - 6 = 1/5x - 2
To eliminate the fraction, multiply both sides of the equation by 5:
5y - 30 = x - 2
Rearrange the equation to put it in the standard form:
x - 5y = 28
Therefore, the equation of the line perpendicular to 5x + y = 57 and passing through the point (10,6) is x - 5y = 28.