Calculate the slope of the secant line through the points on the graph where x=1 and x=3.

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.

To graph this equation, we see that it is in the slope/intercept form of y=mx + b, where m is the slope and b is the y-intercept.

d(t)= -2t+18

Start with the y-intercept of 18 (0,18)

From this point, we can start graphing the line by using the slope of -2. Since the slope is negative, it goes up to the left. So we go one to the left, and up two.

Good luck.

Paul

The easiest way to solve and graph and equation is to put the equation into the slope intercept form y = mx + b, where m is the slope and b in the y-intercept.

To do this, we subtract 4x from both sides and get y = -4x + 0.

From this we know m = -4 (slope) and the y-intercept is 0.

I always start with the y-intercept and put a point there. Thus, we have a point at (0,0). Using this as a starting point, we now use the slope of -4 to get future points. Since it is negative, we go one unit to the left and up four units. So we have the point (-1,4). Using a ruler, we connect these points and continue on both sides to produce the line.

There is also a great free online program/app called Desmos that you can use to check your work. Type in the equation of the line and it will graph it for you.

Good luck,

Paul

The easiest way to solve and graph and equation is to put the equation into the slope intercept form y = mx + b, where m is the slope and b in the y-intercept.

To do this, we subtract 4x from both sides and get y = -4x + 0.

From this we know m = -4 (slope) and the y-intercept is 0.

I always start with the y-intercept and put a point there. Thus, we have a point at (0,0). Using this as a starting point, we now use the slope of -4 to get future points. Since it is negative, we go one unit to the left and up four units. So we have the point (-1,4). Using a ruler, we connect these points and continue on both sides to produce the line.

There is also a great free online program/app called Desmos that you can use to check your work. Type in the equation of the line and it will graph it for you.

Good luck,

Paul

To graph (-3, -3) we start at the origin (0,0) and go 3 squares to the left and then 3 squares down.

Slope is defined in many ways, including rise/run, delta y/ delta x, etc.

Since 0 divided by anything is 0, the rise must be zero, so the slope of 0 is a horizontal line.

Put the point and the slope together, we have a horizontal line going through (-3,-3).

Good luck.

Paul

You cannot. There is technique that can be used to locate a fraction on a line segment. It involves the drawing of parallel lines passing through equidistant points drawn on an intersecting line. The procedure is based on Thales' theorem.
Draw a real line (passing through 0 of course) Mark the limits of segment [0,1]
From the point 0 draw a line. Choose an arbitrary length measured by the opening of a compass. Starting from O, mark three equal segments along the second line. From the end of the 3rd segment draw a line that joins that end with the end of the point 1 on the real line.
From each of the two other points on the secant draw segments parallel to the one you just drew.
Here is how it looks on a picture.
If you have another fraction (5/7) draw 7 equal length segments on the second line (secant to the real axis). The parallel line from the 5th point will cut the real axis at the point 5/7

Select two points P1 and P2 on the line. Read their coordinates P1(X_1, Y_1), and P2(X_2, Y_2). If the points are given to you that is even better: you will not have to approximate their coordinates.
The slope of the straight line that passes through P1 and P2 is the ratio s=(Y_2-Y_1)/(X_2-X_1).
A slope is a number and cannot be graphed. The line that has the slope can be graphed if you have its complete equation y=slope*x + initial value.
To graph a curve on the TI84 open the Y= editor, type in the second member (right side) of the function say on line Y1= then press the Graph function key.

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.

That is an equation describing a straight line. The "slope-intercept" form of a line is

y = mx + b

where m is the slope (change in y-value / change in x-value)
and b is the y-intercept (the point where the line crosses the y-axis when x=0)

Positive slope means the line is rising and negative slope means it's falling.

You can rewrite the original equation 2x - 4y -9 = 0 in slope-intercept form:

y = (1/2)x - (9/4)

So you know the slope is positive 1/2 (line rises 1 y-unit for each 2 x-unit change) and crosses the y-axis at -9/4. With this information you can graph the line.

assuming the question is what is the circle equation?
and if (-2,2) is the center of the circle
the equation should look like this: (x+2)^2+(Y-2)^2=R^2

And now only R is needed.

given 2x-5y+4=0 equation of line perpendicular

we can rearange the equation to be y=(2x+4)/5
from that we can see that the slope of the line is 2/5
And from the fact of perpendicular line we can say that the slope
of the radius line is -2/5.

The motivation now is to calculate the distance between the center of the circle to the cross point of the radius with the line perpendicular

For that we would calculate the radius line equation and compare it to the equation of line perpendicular

As mentioned earlier the slope of the radious line is -2/5.

So the equation is y=-2/5x+b and b can be calculated by using the center of the circle coordinates

2= - (2/5)*(-2)+b ------> b=2-4/5=1.2
radius equation is y=-(2/5)x+1.2

Now the cross point is calculated by comparing the equations:
-(2/5)x+1.2=(2x+4)/5 --> -2x+6=2x+4 --> 4x=2 --> x=1/2 --> y=1

So the cross point is (1/2,1).

The distance between the points is calculated by the following
Formula:

R=SQR(((1/2)-(-2))^2+(2-1)^2)=SQR(2.5^2+1^2)=SQR(6.25+1)=
SQR(7.25)

Therefore the circle eq is (x+2)^2+(Y-2)^2=7.25