Write the equation of a line that is perpendicular to the given line and that passes through the given point.y-3=-1/5(x+2);(-2,7)
To find the equation of a line that is perpendicular to the given line, we need to find the negative reciprocal of the slope of the given line.
The given line has a slope of -1/5. Therefore, the perpendicular line will have a slope of 5 (the negative reciprocal of -1/5).
Now that we have the slope of the perpendicular line, we can use the point-slope form of a linear equation to find the equation of the line. The point-slope form is given by:
y - y₁ = m(x - x₁)
where (x₁, y₁) is the given point and m is the slope of the line.
Plugging in the values:
(x₁, y₁) = (-2, 7)
m = 5
We get:
y - 7 = 5(x - (-2))
Simplifying:
y - 7 = 5(x + 2)
Expanding the expression:
y - 7 = 5x + 10
Moving the constant term to the other side:
y = 5x + 10 + 7
Combining like terms:
y = 5x + 17
So, the equation of the line that is perpendicular to the given line and passes through the point (-2, 7) is:
y = 5x + 17.