Y-y1 = m (x-x1)

The equation of a straight line is given by y=mx + b where y if the y intercept on a graph and x is the x intercept on a graph and m is the slope of the line

Using the given information of ( -4, 6), m=8

-4 is the x intercept, 6 is the y intercept and the slope is 8.

Plugging these values into the y = mx + b gives 6 = (8) (-4) + b solving for b gives b = 32

Going back to the original equation of a straight line gives the equation of y = 8x + 32

Hope this helps Loringh PS Please leave a rating for me Thank you

Posted on Nov 19, 2008

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Posted on Jan 02, 2017

y-y1=((y2-y1)/(x2-x1))(x-x1)

See cap image below

See cap image below

Nov 12, 2013 | Texas Instruments TI-84 Plus Calculator

Touch Screen IC 4182A and MT6301 same. See X- X+ Y+ Y- reference my picture jumper.

**Touch Screen use of 4-wire resistance screen, four lines are: X1, X2, Y1, Y2 **

X1, X2 in the horizontal direction for a change in the resistance, Y1, Y2 for the longitudinal resistivity.

How do you know which line is a two x1 x2 y1 y2? Touch panel along the four lines above are facing their own look at the screen from top to bottom in the line is about on both sides of the x1 x2 is y1 y2. Established phone touch screens measuring four points of the voltage. Two 2.7V and two 0v . two voltage are measured after boot. 2.7 v is y. X is not the voltage.

After corresponding with xy can generally be connected. If the touch panel screen and click position opposite or reverse. Swap contrast x-y reversed.

X1, X2 in the horizontal direction for a change in the resistance, Y1, Y2 for the longitudinal resistivity.

How do you know which line is a two x1 x2 y1 y2? Touch panel along the four lines above are facing their own look at the screen from top to bottom in the line is about on both sides of the x1 x2 is y1 y2. Established phone touch screens measuring four points of the voltage. Two 2.7V and two 0v . two voltage are measured after boot. 2.7 v is y. X is not the voltage.

After corresponding with xy can generally be connected. If the touch panel screen and click position opposite or reverse. Swap contrast x-y reversed.

on Jan 06, 2010 | Cell Phones

m, the slope of a straight line is also called the rate of change of the function. Its definition is **m=(y2-y1)/(x2-x1)** where (x1,y1) are the coordinates of one point on the line, and (x2,y2) are the coordinates of a second point on the lines.

Feb 08, 2012 | Office Equipment & Supplies

Mar 26, 2011 | Borland Turbo C++ 3.0 Full Version...

The site seems to eat the plus signs I enter, so I will use PLUS to symbolize addition.

To find the equation of the straight line (y = a*x PLUS b) that passes through two points P1(x1,y1) and P(x2,y2) , you need to use

1. the coordinates of the points to calculate the slope a (gradient) as a=(y2-y1)/(x2-x1)

2. Replace the calculated value of a in the equation and write that one of the points ( P1(x1,y1) for example) satisfies the equation. In other words y1=a*x1 PLUS b.

Here y1 and x1 are known values, a has been calculated, and only b is still unknown. You can now use the equation y1=a*x1 PLUS b to calculate b as

b=(y1-a*x1)

Example: Equation of the line through (1,5) and (3,6)

Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=6, y1=5, x2=3, and x1=1. You should get a=(6-5)/(3-1)=1/2

The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=(1/2)*3 PLUS b.

Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5

Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

I trust you can substitute you own values for (x1,y1, x2,y2) to duplicate the calculations above.

To find the equation of the straight line (y = a*x PLUS b) that passes through two points P1(x1,y1) and P(x2,y2) , you need to use

1. the coordinates of the points to calculate the slope a (gradient) as a=(y2-y1)/(x2-x1)

2. Replace the calculated value of a in the equation and write that one of the points ( P1(x1,y1) for example) satisfies the equation. In other words y1=a*x1 PLUS b.

Here y1 and x1 are known values, a has been calculated, and only b is still unknown. You can now use the equation y1=a*x1 PLUS b to calculate b as

b=(y1-a*x1)

Example: Equation of the line through (1,5) and (3,6)

Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=6, y1=5, x2=3, and x1=1. You should get a=(6-5)/(3-1)=1/2

The equation is y=(1/2)x PLUS b, where b is not known yet.

To find b, substitute the coordinates of one of the points in the equation. Let us do it for (3,6).

The point (3,6) lies on the line, so 6=(1/2)*3 PLUS b.

Solve for b: 6 MINUS 3/2=b, or b=9/2=4.5

Equation is thus y=(x/2) PLUS 9/2 =(x PLUS 9)/2

I trust you can substitute you own values for (x1,y1, x2,y2) to duplicate the calculations above.

Jan 27, 2011 | Texas Instruments TI-84 Plus Calculator

Calculate the slope (gradient) of the line as a=(y2-y1)/(x2-x1) where y2=1, y1=0, x2=0, and x1=-6. You should get a=(1-0)/(0-(-6))=1/6

The y-intercept is the y-cordinate for x=0. Its value is 1.

The equation is then y=(x/6) 1.

The y-intercept is the y-cordinate for x=0. Its value is 1.

The equation is then y=(x/6) 1.

Oct 18, 2010 | Texas Instruments TI-84 Plus Calculator

Let us start by setting the terminology right: To locate a point on a plane you need two coordinates ( an X-value and a Y-value). If the point is in space, you need a third coordinate which may be called z, but let's us not complicate things unecessarily.

With just two coordinates, we will be able to locate at best one point.

Let us rephrase the problem: When given two sets of coordinates, how to calculate the distance between the two points.

Let (x1,y1) and (x2,y2) be the coordinates of two points in a plane. To calculate the distance between the points, one uses the formula

d (distance)= square root of ( (x2-x1)^2+(y2-y1)^2)

When you calculate the distance you must substitute actual coordinates for X1, Y1, X2 and Y2.

As regards the bearing, I am afraid that I am no expert in maritime nor in aircraft navigation and I will not venture stray out of my area of competence. However, I know that you need an axis that defines the direction with respect to which angles are measured.

If your reference axis is the horizontal axis on a cartesian plane you can determine the angle that the line joining the points makes with that horizontal axis by calculating its cosine, then extract the arcosine.

If (X2-X1) and (Y2-Y1) are both positive then cos(theta)=(X2-X1)/d, where d is the distance (positive value) calculated above.

With just two coordinates, we will be able to locate at best one point.

Let us rephrase the problem: When given two sets of coordinates, how to calculate the distance between the two points.

Let (x1,y1) and (x2,y2) be the coordinates of two points in a plane. To calculate the distance between the points, one uses the formula

d (distance)= square root of ( (x2-x1)^2+(y2-y1)^2)

When you calculate the distance you must substitute actual coordinates for X1, Y1, X2 and Y2.

As regards the bearing, I am afraid that I am no expert in maritime nor in aircraft navigation and I will not venture stray out of my area of competence. However, I know that you need an axis that defines the direction with respect to which angles are measured.

If your reference axis is the horizontal axis on a cartesian plane you can determine the angle that the line joining the points makes with that horizontal axis by calculating its cosine, then extract the arcosine.

If (X2-X1) and (Y2-Y1) are both positive then cos(theta)=(X2-X1)/d, where d is the distance (positive value) calculated above.

Feb 27, 2010 | Casio FX-7400G Plus Calculator

Hello,

In parametric mode you are not drawing one function, but two functions X1(T) and Y1(T). If you use the DrawInv( command either on X1(T) or Y1(T) you get a syntax error.

However if you are drawing functions in Funct mode you can draw the inverse of the function that has been graphed.

Here is how you do it.

[2nd][Draw][8: DrawInv] [ENTER]

The command appears on the home screen, and you need to tell it the function it will act on. To do so press

[VARS] --> Y-Vars [1:Function]

and select the function you want, say Y1, if Y1 has already been defined.

DrawInv Y1 [ENTER]

Both Y1 and its inverse will be drawn.

Hope it helps.

In parametric mode you are not drawing one function, but two functions X1(T) and Y1(T). If you use the DrawInv( command either on X1(T) or Y1(T) you get a syntax error.

However if you are drawing functions in Funct mode you can draw the inverse of the function that has been graphed.

Here is how you do it.

[2nd][Draw][8: DrawInv] [ENTER]

The command appears on the home screen, and you need to tell it the function it will act on. To do so press

[VARS] --> Y-Vars [1:Function]

and select the function you want, say Y1, if Y1 has already been defined.

DrawInv Y1 [ENTER]

Both Y1 and its inverse will be drawn.

Hope it helps.

Sep 02, 2009 | Texas Instruments TI-84 Plus Silver...

It depends on how complex the function is; there are different techniques. For a straight line of the standard form, usually shown as y=mx+b; this is a simple function of y in terms of x. So, determining the function in this case means finding values for m and b. --- m is the slope of the line, commonly called rise over run. Since your question asks in terms of coordinates, I would assume that you are looking at a problem giving you 2 points.

Probably something like p1 = (3,4); p2 = (5,8); the general form of this is p1=(x1,y1); p2=(x2,y2). The slope in this case is rise over run, in other words, the change in y divided by the change in x. This can be calculate between these two points as (y2-y1)/(x2-x1). In my sample above, this would be (8-4)/(5-3) = 4/2 = 2 giving the value of m.

To find the value of b (the y intercept) you need the value of y when x = 0. Since you already know that m, the slope is 2, consider a new point, call it p3 (x3,y3), pick either of the known points and solve the slope equation again, this time for y3. [m=(y3-y1)/(x3-x1)]. We know that x3 is 0, since we are trying to solve for y where that is true, so the equation becomes:

m=(y3-y1)/(-x1)

-x1*m=y3-y1

y3=y1-x1*m

y3 is really b in the standard form, since it is, by definition the intercept, or the value when x=0,

so

b=y1-x1*m -- this gives you the y intercept anytime you know the slope and one point on the line.

In the example, y1 = 4, x1=3 and we've already calculated m to be 2, so

b=4-3*2 = -2

So, the function would be

y = 2x -2

To check, plug in the values of the other point, p2, and see if they work

y2 = 2*x2 - 2

8=2*5-2

It's easier than it looks. It can help you to understand if you get some old fashioned graph paper and plot it so you can see what is happening.

Probably something like p1 = (3,4); p2 = (5,8); the general form of this is p1=(x1,y1); p2=(x2,y2). The slope in this case is rise over run, in other words, the change in y divided by the change in x. This can be calculate between these two points as (y2-y1)/(x2-x1). In my sample above, this would be (8-4)/(5-3) = 4/2 = 2 giving the value of m.

To find the value of b (the y intercept) you need the value of y when x = 0. Since you already know that m, the slope is 2, consider a new point, call it p3 (x3,y3), pick either of the known points and solve the slope equation again, this time for y3. [m=(y3-y1)/(x3-x1)]. We know that x3 is 0, since we are trying to solve for y where that is true, so the equation becomes:

m=(y3-y1)/(-x1)

-x1*m=y3-y1

y3=y1-x1*m

y3 is really b in the standard form, since it is, by definition the intercept, or the value when x=0,

so

b=y1-x1*m -- this gives you the y intercept anytime you know the slope and one point on the line.

In the example, y1 = 4, x1=3 and we've already calculated m to be 2, so

b=4-3*2 = -2

So, the function would be

y = 2x -2

To check, plug in the values of the other point, p2, and see if they work

y2 = 2*x2 - 2

8=2*5-2

It's easier than it looks. It can help you to understand if you get some old fashioned graph paper and plot it so you can see what is happening.

Jun 16, 2009 | Computers & Internet

I'm sure there are easier ways to do this but the "brute force" method is to label point A as x1, y1; point b as x2, y2 and point c as x3,y3. Then just do square root of the sum of: (x2-x1)squared+(y2-y1)squared. Save this into memory 'A'. Now do the same for x2,y2 to x3,y3 and add this to the number saved in 'A' and do it once more for x3, y3 to x1, y1 and add to the number in 'A' to get the result.

Nov 08, 2008 | Office Equipment & Supplies

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