write an equation of the line that is perpendicular to the given line the passes through the given point .
A. y=- 3x + 7 passing through (3, 5)
To find the equation of a line that is perpendicular to the given line, we first need to determine the slope of the given line.
The given line has the equation y = -3x + 7, where the coefficient of x (-3) represents the slope.
A line perpendicular to this line will have a slope that is the negative reciprocal of -3. The negative reciprocal of -3 is 1/3.
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 passing through the given point (3, 5):
y - y₁ = m(x - x₁),
where (x₁, y₁) is the given point, and m is the slope of the line.
Plugging in the values, we get:
y - 5 = (1/3)(x - 3).
Rearranging the equation and distributing the (1/3), we have:
y - 5 = (1/3)x - 1.
To make it more standard, we can multiply through by 3 to eliminate the fraction:
3(y - 5) = 3(1/3)x - 3(1).
Simplifying, we get:
3y - 15 = x - 3.
Finally, rearranging the equation to isolate y, we have the equation of the line perpendicular to y = -3x + 7 that passes through (3, 5):
3y = x + 12.