Analytic geometry (-2,2) 2x-5y+4=0 equation of line perpendicular

that sounds like a problem about "slopes" so I am looking at using the derivative, which is basically the slope of the graph.
- the equation of line PQ can be simplified to y(x) = -2kx + k -8
- derivative relative to X (k is a constant) gives you -2k
- derivative of y(x) = 4x + 7 relative to x gives you 4

if the lines are parallel then the slopes are equal:
-2k=4 gives you k = -2 => your function becomes y = 4x - 8

if the lines are perpendicular the two slopes multiplied together give you -1:
-2k * 4 = -1 gives you k=1/8 => function becomes y = -1/4x - 7/8 or in other words y = -(2x+7)/8

I have made graphic representation and it looks to be correct.
I hope it makes sense.

Let's break this down into a few parts first.

Slope intercept form is otherwise known as y = mx + b, where m is the slope and b is the y-intercept.
Perpendicular means that the two lines will intersect each other at a 90 degree angle. In math terms, their slopes will be negative reciprocals of each other.
Starting with y + 4 = 1/4(x -7), putting into slope intercept form, y = mx + b, m = 1/4 and b=-7/4 -4.

Since it is perpendicular, it must have a slope that is a negative reciprocal. The slope of the original line is 1/4, so the multiplying it -1, we get -1/4, and taking the reciprocal, we get -4. So the slope of the perpendicular line is -4.

So, y = -4x + b, but we don't know what the value of b is. To determine this, we know the point (-4,7) is on the line that we are trying to find, so we can substitute it into the equation and calculate b to make it work.

Every time we see an x we put in -4 and every time we see a y, we put in 7.

y = -4x + b
7 = -4(-4) + b
7 = 16 + b
subtract 16 from both sides
7 - 16 = 16 + b - 16
-9 = b

Now substitute this into the equation.

y = -4x + -9
Putting it into correct form, we get y = -4x - 9.

Let's check it to see if it is correct.

It has a slope of -4, so it is perpendicular to y=1/4x - 5 3/4.
Is the point (-4,7) on the line? Let's substitute it in to see if it is on the line.

Again, everywhere we see an x, we put in -4 and everywhere we see a y we put in 7.

y= -4x - 9
7 = -4 (-4) - 9
7 = 16 - 9
7 = 7

Sorry for the very long explanation, I was being overly thorough, but after you do a few of them, you will be able to knock them off in minutes.

Good luck,

Paul

This is best written as two separate equations:

8x+3y = -23 and 34x+ 5y = -23
Solving the first one for x:
8x = -23-3y
x = -23/8 - 3/8y
Substituting this value for x into the second equation:
34(-23/8 - 3/8y) + 5y = -23
-97.75 - (34)(.375)y + 5y = -23
-97.75 - 12.75y + 5y = -23
-97.75 -7.75y = -23
-7.75y = 97.75-23=74.75
y = -74.75/7.75 = -9.645161
Substitution back into the equation for x:
x = -23/8 - 3/8(-9.645161)
x = -2.875 + 3.616935
x =.741935

9x-5y=121

First of all you equation is not one : it has nothing on the right side of the = sign. But to answer the general question let us write the equation as 50+25x-5y=0
X-intercept (also know as roots) There may be several
Definition: X-intercepts are those values of the independent variable x for which y=0. For a straight line there con be at most 1 x-intercept.
To find the intercept, set y=0 in the equation of the line and solve for x
50+25x-5(0)=0 or 50+25x=0. The solution is x=-(50/25)=-2

Y-intercept (also know as the initial value. There can only be 1 y-intercept, otherwise the expression does not represent a function.
Definition: It is the value of the dependent variable y when x=0 (where the function crosses the y-axis
To find it, set the x-value to 0 in the equation of the line.
50+25x-5y=0
50+25(0)-5y=0, or 50-5y=0. The solution is y=50/5=10
The straight line cuts the x-axis at the point (-2, 0) and the y-axis at the point (0,10)

Two lines are perpendicular if they belong to the same plane and intersect (cut) one another at right angle. They make a 90 degree angle. If you are doing analytic geometry, two lines are perpendicular if the product of their slopes ia equal to -1.
Other relative positions of lines are parallel (they have the same slope/direction) or they are just secant at an angle not equal to 90 degrees.

Calcualte the slope of the line as
a=(7-(-11))/(10-(-5))=18/15=6/5
Use the fact that the line passes through one of the two points, for example (10,7)
7=(6/5)*10+b=12+b
Obtain b as b=7-12=-5
The equation of the line in functional form is y=(6/5)x-5
Multiply everything by 5 to clear the fraction
5y=6x-25 or 0=6x-5y-25
Finally, the equation in general form (standard?) is 6x-5y-25=0.

Check the calculation by verifying that the point (10,7) lies on the line.
6(10)-5(7)-25=60-35-25=60-60=0 CHECKed!
Check that the second point (-5,-11) lies on the line also (if you want to)
6*(-5)-5*(-11)-25=-30+55-25=0
That checks OK.

x-intercept (or zero of the function)
Set y=0 in the equation and solve fo x: 2x-5(0)=20 and x=20/2=10

y-intercept (initial value)
Set x= 0 in the equation and solve for y: 2(0)-5y=20 and y=20/(-5)=-4

If the equation is in the slope-intercept form y=ax+b
1. y-intercept is b
2. To get the x-intercept ( zero of the function) Set y=0=ax+b, and x=-b/a

Hello,
And they are perpendicular. But due to the finite resolution of the screen they do not appear to be. However if you had a geometry application linked with the function graphing the measurement of the angle between the lines will show that it is 90 degrees, or 1.58 rad.
This screen capture from the TI'Nspire will convince you, I hope.

Hope it helps.

nice..let's see..counts on fingers..

x=a(b)-(d-n)/z.com