write an equation for a line perpendicular to y=-5x+2 and passing through the point (5,2)
To find the equation of a line that is perpendicular to y = -5x + 2 and passes through the point (5,2), we can first determine the slope of the given line and then determine the negative reciprocal of that slope.
The given line y = -5x + 2 is in slope-intercept form (y = mx + b), where m is the slope. Comparing it with the standard form, we can see that its slope is -5.
To find the negative reciprocal of -5, we invert the fraction and change the sign:
Negative reciprocal of -5 = (-1/5)
The slope of the line perpendicular to y = -5x + 2 will have a slope of (-1/5).
Now, we have the slope and the point (5,2) through which the line passes. We can use the point-slope form of a linear equation to find the equation of the perpendicular line:
y - y₁ = m(x - x₁)
where (x₁, y₁) represents the given point and m represents the slope of the perpendicular line.
Substituting the values, we have:
y - 2 = (-1/5)(x - 5)
Expanding and simplifying:
y - 2 = (-1/5)x + 1
Adding 2 to both sides to isolate y:
y = (-1/5)x + 3
So, the equation of the line perpendicular to y = -5x + 2 and passing through the point (5,2) is y = (-1/5)x + 3.