What is the equation of a line that satisfies the given conditions: perpendicular to #y= -2x + 5# and passing through (4, -10)?

Question

Autumn Armstrong · Accepted Answer

#y=0.5x-12#

Since the line must be perpendicular, the slope #m# should be the opposite and inverse of the one in your original function.

#m=-(-1/2)=1/2=0.5#

Now all you have to do is use the point slope equation:

Given coordinate: #(4,-10)#

#y-y_0=m(x-x_0)#

#y-(-10)=0.5(x-4)#

#y+10=0.5x-2#

#y=0.5x-2-10#

#y=0.5x-12#

Evelyn Arboleda · Answer

undefined To find the equation of a line perpendicular to $ y = -2x + 5 $ and passing through the point $ (4, -10) $, we use the fact that the slopes of perpendicular lines are negative reciprocals of each other.

The given line has a slope of $ -2 $. The slope of the perpendicular line will be the negative reciprocal of $ -2 $, which is $ \frac{1}{2} $.

Using the point-slope form of a linear equation $ y - y_1 = m(x - x_1) $, where $ m $ is the slope and $ (x_1, y_1) $ is a point on the line:

Substitute $ m = \frac{1}{2} $ and $ (x_1, y_1) = (4, -10) $:

$ y - (-10) = \frac{1}{2}(x - 4) $

Simplify:

$ y + 10 = \frac{1}{2}x - 2 $

Subtract 10 from both sides:

$ y = \frac{1}{2}x - 12 $

So, the equation of the line perpendicular to $ y = -2x + 5 $ and passing through $ (4, -10) $ is $ y = \frac{1}{2}x - 12 $.

Evelyn Arboleda · Answer

undefined The equation of the line that satisfies the given conditions, perpendicular to $ y = -2x + 5 $ and passing through $ (4, -10) $, is $ y = \frac{1}{2}x - 12 $.