Recall that perpendicular slopes have opposite reciprocals. So when writing the equation of the new line you will need the slope that is perpendicular to the one in y = -2x + 5. Recall that the slope of this line comes from the form y = mx + b where m is the slope. The m is the coefficient (number) in front of x. The slope of the given line then is m = -2. The slope of the perpendicular line would be the opposite reciprocal of -2, which is 1/2.
Then you use the equation for point-slope linear lines y - y1 = m(x - x1) and insert the perp slope m = 1/2 and the point given (3,-1) in for (x1, y1).
The resulting equation would be y + 1 = 1/2(x -3). If they do not ask for the form of the equation, you may leave it in point-slope. They often ask for the form y = mx +b which would mean you have to solve the equation for y:
y + 1 = 1/2(x - 3) by distributing first:
y + 1 = 1/2 x - 3/2 then moving the 1 to the other side by subtracting:
y = 1/2x -3/2 -1 and combining like terms -3/2 -1 and getting:
y = 1/2 x = 5/2
Let us say the equation of the line perpendicular to y=-2x+5 to be y = ax+b where a and b are unknown that should be discovered.
It is given that our line y = ax+b is perpendicular to y = -2x+5.
So the slope of the two lines must be negative reciprocals of each other.
In other words (-2) = -1/(a)
Another way to say this is (-2)(a) = -1 .
(-2)xx(a) = -1
a = (-1)/(-2) = 0.5
So now we can write our line as y = 0.5x+b
It is given that this line passes through (3,-1).
y = 0.5x+b
-1 = 0.5xx3+b
b = -1-1.5 = -2.5
y = 0.5x-2.5
So the answer is y = 0.5x-2.5
Or else we can multiply both sides by 2 and say it as 2y = x-5.
http://www.mesacc.edu/~scotz47781/mat150/notes/eqn_line/Equation_Line_Parallel_Perpendicular_Notes.pdf
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