Which equation is the equation of a line that passes through (-10,3) and is perpendicular to y=5x-7?
1- y=5x+53
2- y=-1/5x -7
3- y= -1/5x +1
4- y= 1/5x +5
(-10,3), m = -1/5.
Y = mx + b
Y = 3
m = -1/5
x = -10
Solve for b(y-intercept).
To find the equation of a line that passes through a given point and is perpendicular to a given line, we need to use the concept of slope.
First, let's find the slope of the given line y = 5x - 7. The given line is in slope-intercept form (y = mx + b), where m represents the slope. So, the slope of this line is 5.
For a line to be perpendicular to another line, the slopes of the two lines must be negative reciprocals of each other. The negative reciprocal of 5 is -1/5.
Now that we have the slope of the desired line (-1/5) and the point it passes through (-10, 3), we can use the point-slope form of a linear equation to find the equation of the line.
The point-slope form of a linear equation is given by:
y - y1 = m(x - x1)
Substituting the values, we have:
y - 3 = (-1/5)(x - (-10))
Simplifying further:
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:
3- y = -1/5x + 1.
So, option 3 is the correct answer:
3- y = -1/5x + 1.