Ok, so you are given an equation in standard form, 5x+4y=2, and a point, (7, 5). You are being asked to write the equation for a line that is parallel to the equation in standard from, and that includes the point (7, 5).
First, let's start by finding the slope of your new line. We know that it needs to have the same slope as 5x+4y=2, because parallel lines have the same slopes. To do that, we need to put the equation into slope-intercept form (y=mx+b), which means we need to isolate the "y."
5x+4y=2
-5x -5x (subtract 5x from both sides to move it to the right side of your equation)
4y = 2 - 5x
/4 /4 /4 (divide all the terms by 4 to get "y" by itself)
y = (1/2) - (5/4)x ... I suggest leaving your slope as a fraction.
Now, we know that our slope, m , is going to be -(5/4).
Next, we are going to use our slope, -(5/4), and point, (7, 5), to find the b-value (y-intercept) of your new line. Let's plug in what we know:
y = 5
m = -(5/4)
x = 7
y=mx+b
5=(-5/4)(7) + b -> I plugged in what we knew for y, m, and x.
5 = -(35/4) + b -> I multiplied the numbers in the numerator (5 x 7) to get 35/4
20/4 = -(35/4) + b -> I converted 5 into a faction with a denominator of 4 by multiplying by (4/4)
+(35/4) +(35/4) -> I add (35/4) to both sides to isolate b
55/4 = b ... or b = 13.74
Answer:
m = -(5/4)
b = 13.75
The equation perpendicular to the line going through the points (5, -1) and (7, -5) is [tex]y=\frac{1}{2} x-\frac{7}{2}[/tex]
The line passes through the points (5, -1) and (7, -5)
That is [tex]x_1=5, y_1=-1, x_2=7, y_2=-5[/tex]
Calculate the slope(m) using the formula below:
[tex]m=\frac{y_2-y_1}{x_2-x_1} \\\\m=\frac{-5-(-1)}{7-5} \\\\m=\frac{-4}{2}\\\\m=-2[/tex]
The equation of the new line perpendicular to the line above will be given as:
[tex]y-y_1=\frac{-1}{m} (x-x_1)\\\\y-(-1)=\frac{-1}{-2} (x-5)\\\\y+1=\frac{1}{2}(x-5)\\\\y=\frac{1}{2} x-\frac{5}{2} -1\\\\y=\frac{1}{2} x-\frac{7}{2}[/tex]
The equation perpendicular to the line going through the points (5, -1) and (7, -5) is [tex]y=\frac{1}{2} x-\frac{7}{2}[/tex]
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