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Posted on Jan 02, 2017

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SOURCE: Find an equation of the line containing the given

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.

Posted on Oct 19, 2010

SOURCE: i knowi can find a

The general answer is no. However the calculator can be used to find the slope and the y-intercept. You write the formula for the rate of change (a) and use the calculator to calculate it. Then you use the formula to find the y-intercept.

If you have 2 points, you can use the 2 -var statistics program to find a linear regression.

Posted on Jul 06, 2011

5x + 7 = 45

5x = 45 -7

5x = 38

X = 38/5 = 7 3/5 or in decimal

X = 7.6

Or this could be a straight line equation of y = mx + c

Were y = 45 m = 5 = gradient of 5. Through the point of origin c = 7 at x =0

Question not specific enough

What topic are you doing?

5x = 45 -7

5x = 38

X = 38/5 = 7 3/5 or in decimal

X = 7.6

Or this could be a straight line equation of y = mx + c

Were y = 45 m = 5 = gradient of 5. Through the point of origin c = 7 at x =0

Question not specific enough

What topic are you doing?

Feb 26, 2018 | Homework

This means the line passes through (0,4) and (2,1)

So look here

http://www.mathsisfun.com/algebra/line-equation-2points.html

and plug in these values to get

(y - 4) / (x - 0) = (1 - 4) / (2 - 0)

y - 4 = - 1.5 x or y = -1.5 x + 4

So look here

http://www.mathsisfun.com/algebra/line-equation-2points.html

and plug in these values to get

(y - 4) / (x - 0) = (1 - 4) / (2 - 0)

y - 4 = - 1.5 x or y = -1.5 x + 4

Nov 05, 2014 | Office Equipment & Supplies

Assuming the 'standard form' is "slope-intercept", calculate the slope from the equation m = __y2-y1__ =__ 5 - 1__ = __ 4__ = -2

x2-x1 4 - 6 -2

The intercept can be found by substituting either of the two points into the equation y = mx + b

5 = (-2)4 + b

5 = (-8) + b

13 = b

(OR, using the other point, y = mx + b

1 = (-2)6 + b

1 = (-12) + b

13 = b )

Then expressing in general:

**y = (-2) x + 13**

x2-x1 4 - 6 -2

The intercept can be found by substituting either of the two points into the equation y = mx + b

5 = (-2)4 + b

5 = (-8) + b

13 = b

(OR, using the other point, y = mx + b

1 = (-2)6 + b

1 = (-12) + b

13 = b )

Then expressing in general:

Oct 10, 2014 | Computers & Internet

Being parallel to the given line, the equation of the line you are seeking has the same slope, which in this case is **a=1/4.**

So the equation sought is as follows

y=**(1/4)x** +b, where b is to be found.

To find** b**, use the stated fact that the line passes through the point **(x=8, y=-1)**. All that means is that the point **(8,-1)** is on the line whose equation you are looking for. If it is on the line with equation **y=(1/4)x+b**

then its coordinates x=8, and y=-1 satisfy the relation y=(1/4)x+b. In other words, if you substitute 8 for x, and -1 for y, the equality holds true**-1=(1/4)*8 +b**

This gives you a way to find the initial value of the function (the y-intercept b ). Just solve**-1=(1/4)*8 +b** to find b.

I leave this pleasure to you.

So the equation sought is as follows

y=

To find

then its coordinates x=8, and y=-1 satisfy the relation y=(1/4)x+b. In other words, if you substitute 8 for x, and -1 for y, the equality holds true

This gives you a way to find the initial value of the function (the y-intercept b ). Just solve

I leave this pleasure to you.

Jan 15, 2014 | Mathsoft StudyWorks! Mathematics Deluxe...

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

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=3/2 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

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=3/2 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

Jan 10, 2011 | Texas Instruments TI-84 Plus Calculator

What you call the gradient is usually called the slope of the line. Select two points P1 and P2 on the line. Read their coordinates P1(X_1, Y_1), and P2(X_2, Y_2).

The slope of the straight line that passes through P1 and P2 is the ratio s=(Y_2-Y_1)/(X_2-X_1).

The slope of the straight line that passes through P1 and P2 is the ratio s=(Y_2-Y_1)/(X_2-X_1).

Nov 18, 2010 | Casio fx-300ES Calculator

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=3/2 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

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=3/2 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

Oct 20, 2010 | 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

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

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

Oct 26, 2008 | Casio FX-115ES Scientific Calculator

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