last one
write an equation of the line that passess through 10,-4 and is perpendicular to the line whose equation is y=2/5x+8
help
you want a line with slope -5/2.
Now you have a point and a slope. By now that should make writing the line's equation a cinch.
please help with my other ones
To find an equation of a line that is perpendicular to another line, you need to use the slope of the given line. In this case, the equation of the given line is y = (2/5)x + 8.
The given line has a slope of 2/5. To find the slope of a line perpendicular to it, you need to calculate the negative reciprocal of this slope. The negative reciprocal of 2/5 is -5/2.
Now that you have the perpendicular slope, you can use the point-slope form of a linear equation to find the equation of the line passing through the point (10, -4). The point-slope form is:
y - y1 = m(x - x1)
where (x1, y1) is the point that the line passes through, and m is the slope.
Plugging in the values, we have:
y - (-4) = (-5/2)(x - 10)
Simplifying this equation, we get:
y + 4 = (-5/2)(x - 10)
Now, let's expand and simplify further:
y + 4 = (-5/2)x + 25
Subtracting 4 from both sides, we have:
y = (-5/2)x + 21
Therefore, the equation of the line that passes through (10, -4) and is perpendicular to y = (2/5)x + 8 is y = (-5/2)x + 21.