Find the line of the equation that
hits y=4x^2-20 when x= -5 that is
perpendicular to 5x-6=3y
My solution does not graphical show perpendicular lines, but algebraically it works lol Help!
when x=-5
y = 4(-5)^2 - 20 = 80
so you have the point (-5,80)
the slope of 5x-6 = 3y is 5/3
so the slope of the required line is -3/5
Now use whichever method you normally use to find the equation of a line given its slope and a point on it.
Hi Reiny, Yes, I got that far. I even came up with a line, but am looking to see what others get as my line, when graphed, is not perpendicular to the other line mentioned in the prob (5x-6=3y)
since the new line has slope -3/5, its equation must be
3x + 5y = k
point(-5,80) on it ....
-15 + 400 = k = 385
so new equation is 3x + 5y = 385 or
y = (-3/5)x + 77
Since the slopes are negative reciprocals of each other, their graphs MUST be perpendicular.
Check your calculation of points and check the plotting of points
I hope you're right since that's the equation I got too! I may be entering things into my calculator wrong. Using that is new to me. Thanks for all you're help!
To find a line that is perpendicular to another line, we need to use the slope-intercept form of the equation, which is y = mx + b, where m represents the slope. In order for two lines to be perpendicular, their slopes must be negative reciprocals of each other.
Let's first find the slope of the given line, 5x - 6 = 3y. To do this, we need to rewrite the equation in slope-intercept form.
5x - 6 = 3y
3y = 5x - 6
y = (5/3)x - 2
Comparing this equation to y = mx + b, we can determine that the slope of the given line is m = 5/3.
To find a line perpendicular to this, we need to find the negative reciprocal of 5/3, which is -3/5.
So, the slope of the line we are looking for is m = -3/5.
Now, we have the slope and a point on the line (x = -5 and y = 4x^2 - 20). We can use the point-slope form of the equation to find the equation of the line.
The point-slope form of the equation is: y - y1 = m(x - x1).
Substituting the given values: y - (-20) = (-3/5)(x - (-5)).
Simplifying this equation, we get:
y + 20 = (-3/5)(x + 5).
Further simplifying, we have:
y + 20 = (-3/5)x - 3.
To put this equation in slope-intercept form, we isolate y:
y = (-3/5)x - 3 - 20
y = (-3/5)x - 23.
Therefore, the equation of the line that is perpendicular to 5x - 6 = 3y and passes through the point (x = -5, y = 4x^2 - 20) is y = (-3/5)x - 23.