Which equation is the equation of a line that passes through (-10,3) and is perpendicular to y=5x-7?
y-3 = -1/5 (x+10)
find the equation of the line,in slope-intercept form with a slope of -8 and passes through the -2,2
To find the equation of a line that is perpendicular to another line, we need to determine the slope of the given line and then find the negative reciprocal of it. The equation of a line in slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept.
The equation y = 5x - 7 is in slope-intercept form. Comparing it with y = mx + b, we can see that the slope of the given line is 5.
To find the negative reciprocal of the slope, we invert it and change its sign. Thus, the negative reciprocal of 5 is -1/5.
Since the new line is perpendicular to the given line and passes through the point (-10, 3), we can use the point-slope form of a linear equation: y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line.
Plugging in the values, we have:
y - 3 = (-1/5)(x - (-10))
Next, simplify and rearrange the equation:
y - 3 = (-1/5)(x + 10)
y - 3 = (-1/5)x - 2
y = (-1/5)x + 1
Therefore, the equation of the line that passes through (-10, 3) and is perpendicular to y = 5x - 7 is y = (-1/5)x + 1.