To find the equation of a line that is perpendicular to the given line, we first need to find the slope of the given line. The given line is in the form y - 3 = -(x + 2), which can be rewritten in slope-intercept form as y = -x - 5. The slope of this line is -1.
The slope of a line perpendicular to another line is the negative reciprocal of the slope of the original line. Therefore, our perpendicular line will have a slope of 1.
We also know that the perpendicular line passes through the point (-2, 7).
Using the slope-intercept form of a line (y = mx + b), where m is the slope and b is the y-intercept, we can substitute the slope (1) and the given point (-2, 7) into the equation:
y = mx + b
7 = 1(-2) + b
7 = -2 + b
Solving for b, we find that b = 9.
Therefore, the equation of the line that is perpendicular to y - 3 = -(x + 2) and passes through (-2, 7) is y = x + 9.
Write the equation of a line that is perpendicular to the given line and that passes through the given point.
1 y-3= -(x+2); (-2,7)
(1 point)
y = 5x +7
y = 5x + 17
1 Oy=3x-2
y=-2x+3
1 answer