perpendicular to line -2x+y=7 contains the point (-4,-2) the equation of line is
-2x+y=7 can be written as
y=2x+7, so we see it has slope=2
so, the perpendicular has slope -1/2
now you have a point and a slope:
y+2 = -1/2 (x+4)
and you can massage that as you will.
6x+2y=12
find intercept
a circle equation x^2+y^2+4x-4y-1=0 and graph
2)find domain and range f(x)=x^2-4x-5
find vertex of f(x) =x^2+4x-21
To find the equation of a line that is perpendicular to another line and passes through a given point, you need to follow these steps:
Step 1: Identify the slope of the given line.
The given line has the equation -2x + y = 7. To find its slope, we need to rearrange the equation into slope-intercept form (y = mx + b), where m is the slope. So, let's rearrange the equation:
y = 2x + 7
From this equation, we can see that the slope of the given line is 2.
Step 2: Determine the negative reciprocal of the slope.
Since we want to find a line that is perpendicular to the given line, we need to determine the negative reciprocal of the slope. The negative reciprocal of 2 is -1/2.
Step 3: Write the equation of the perpendicular line using the given point and the negative reciprocal slope.
The given point is (-4, -2). Let's use the point-slope form of a line to write the equation:
y - y1 = m(x - x1)
where (x1, y1) is the given point and m is the negative reciprocal slope.
Substituting the values into the equation, we have:
y - (-2) = -1/2(x - (-4))
y + 2 = -1/2(x + 4)
y + 2 = -1/2x - 2
y = -1/2x - 4
So, the equation of the line that is perpendicular to -2x + y = 7 and passes through the point (-4, -2) is y = -1/2x - 4.