the equation of a line perpendicular to the line y = x - 3 and containing the point (8, -2).
slope of given line = 1
slope of perpendicular line = -1
so new equation :
y = -x + b
plug in the point ...
-2 = -1(8) + b
b = 6
new equation: y = -x + 6
To find the equation of a line that is perpendicular to another line, we need to consider the slopes of the two lines.
The given line is in the form y = mx + b, where m represents the slope. In this case, the slope of the line y = x - 3 is 1.
For a line to be perpendicular to another line, the product of their slopes should be -1 (since the product of perpendicular slopes is always -1).
So, let's find the slope of the line we are looking for:
The slope of the given line is 1. To find the perpendicular slope, we can take the negative reciprocal of 1.
The negative reciprocal of 1 is -1/1, which is just -1.
Now that we know the slope of the line we are looking for is -1, we can use the point-slope form of a line to find the equation. The point-slope form is y - y1 = m(x - x1), where (x1, y1) represents the coordinates of a point on the line and m represents the slope.
We are given the point (8, -2) that the line should pass through, so we can substitute these values into the point-slope form:
y - (-2) = -1(x - 8)
Simplifying this equation, we get:
y + 2 = -x + 8
Rearranging the equation to the slope-intercept form, which is y = mx + b, where b represents the y-intercept, we get:
y = -x + 6
So, the equation of the line perpendicular to y = x - 3 and containing the point (8, -2) is y = -x + 6.