Write in y=mx+b form the line perpendicular to 5x-y=2 passing through (10,-2)
Solve for y for the given equations. Putting it in y=mx+b form. Then do the negative reciprocal of the slope from the given equation.
After that use the point slope formula to calculate the new equation using the given point and your new slope.
Do you get it?
I got -y=-5x+2 but I am confused on the actual math part. To get rid of the - on the y do I do y=1/5x+1/2?
Oh, you just divide the negative to each side and get y=5x-2.
Then use y-y1=m(x-x1)
y-(-2)=-1/5(x-10)
Then solve.
Oh and the answer would be y=-1/5x
so just to check is the answer y=-1/5x? Thanks for all your help by the way.
Ha Ha Thanks I didn't see the next post. Thanks so much you were so much help.
To write the equation of a line perpendicular to another line, we need to find the slope of the given line and then find the negative reciprocal of that slope.
First, let's rewrite the given equation in the standard form ax + by = c, where "a," "b," and "c" are integers.
Given equation: 5x - y = 2
To convert it to standard form, we need to isolate the "y" term:
y = 5x - 2
Now we can see that the coefficient "m" of the x-term is 5, which represents the slope of the given line. To find the slope of the line perpendicular to it, we take the negative reciprocal of 5:
Negative reciprocal of 5: -1/5
Now we have the slope of the perpendicular line, -1/5.
Next, we use the point (10, -2) to find the y-intercept, "b," of the perpendicular line.
We can substitute the coordinates (x, y) = (10, -2) into the slope-intercept form, y = mx + b, and then solve for "b."
-2 = (-1/5)(10) + b
-2 = -2 + b
b = -2 + 2
b = 0
The y-intercept, b, is 0.
Now, we have the slope, m = -1/5, and the y-intercept, b = 0, for the perpendicular line.
Thus, the equation of the line perpendicular to 5x - y = 2 and passing through the point (10, -2) can be written in slope-intercept form (y = mx + b) as:
y = (-1/5)x + 0
Simplifying further, the equation becomes:
y = -1/5x