Final answer:
The equation of the line perpendicular to y=x+27 and passing through the point (-3,11) is y=-x+8, derived using the point-slope form and the negative reciprocal slope concept.
Step-by-step explanation:
To write an equation that is perpendicular to y=x+27 and passes through the point (-3,11), first we need to determine the slope of the given line. The equation y=x+27 is in slope-intercept form, y=mx+b, where m is the slope and b is the y-intercept. For the given equation, the slope (m) is 1. Perpendicular lines have slopes that are negative reciprocals of each other. Therefore, the slope of the line perpendicular to y=x+27 will be -1 (the negative reciprocal of 1).
Now, using the slope -1 and the point (-3,11), we can use the point-slope form of the equation of a line, which is y - y1 = m(x - x1), where (x1, y1) is the point the line passes through, and m is the slope, to write the equation of our new line. Substituting the values, we get y - 11 = -1(x + 3).
Next, we'll simplify the equation:
y - 11 = -1(x + 3)
y - 11 = -x - 3
y = -x + 8
The equation of the line that is perpendicular to y=x+27 and passes through the point (-3,11) is y = -x + 8, which is in slope-intercept form (y=mx+b).