"A line perpendicular to x + 3y = 5 passes through (1, -1). What is the
equation of the line in standard form?"
You have a point (1,-1) and a slope 3
The point-slope form is
y-(-1) = 3(x-1)
Rearrange terms to get to standard form:
y+1 = 3x-3
-3x + y = -4
To find the equation of a line that is perpendicular to the given line and passes through the point (1, -1), we can follow these steps:
Step 1: Rewrite the given equation in the slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept.
Given equation: x + 3y = 5
Rearrange the equation to solve for y:
3y = -x + 5
y = (-1/3)x + 5/3
From the equation, we can see that the slope of the given line is -1/3.
Step 2: Determine the slope of the line perpendicular to the given line.
The slope of a line perpendicular to another line is the negative reciprocal of the slope of that line.
So, the slope of the perpendicular line would be the negative reciprocal of -1/3, which is 3.
Step 3: Use the slope-intercept form (y = mx + b) and the given point (1, -1) to find the equation of the perpendicular line.
Substitute the values of the slope (m = 3) and the coordinates of the point (x = 1, y = -1) into the slope-intercept form:
y = mx + b
-1 = 3(1) + b
-1 = 3 + b
b = -4
So, the y-intercept (b) of the perpendicular line is -4.
The equation of the line in slope-intercept form is y = 3x - 4.
Step 4: Convert the slope-intercept form into standard form (Ax + By = C).
To convert the slope-intercept form to standard form, move all terms to the left side of the equation:
-3x + y = -4
Multiply the equation by -1 to make the coefficient of the x- term positive:
3x - y = 4
So, the equation of the line in standard form that is perpendicular to x + 3y = 5 and passes through (1, -1) is 3x - y = 4.