Question about Texas Instruments TI-84 Plus Calculator

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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.

Posted on Jan 27, 2011

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Posted on Jan 02, 2017

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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

Both equations are straight lines with different slopes and different intercepts. Where these two lines cross is the ordered pair (x,y) required. This point is (0.15, 0.115)

Dec 08, 2013 | Mathsoft StudyWorks! Mathematics Deluxe...

**No solutions:**The system is incoherent, incompatible Example: 2x+3y=8 and 2x+3y= 15. The two lines are parallel and distinct.**One solution:**There exits a pair of values (x,y) that satisfy both linear equations. The two lines on a Cartesian graph have one intersection point.**Infinite number of solutions:**The two equations are one and the same (one is just multiplied by some constant). The graph of the two lines yields the same line. One is superposed on the other. Any ordered pair (x,y) that satify one equation (there is an infinity of such pairs) satisfies the other.**Two solutions: cannot happen**because the two lines can either intersect once, be parallel, or superposed one on the other.

Dec 13, 2012 | Mathsoft StudyWorks! Mathematics Deluxe...

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.

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

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.

Dec 04, 2011 | Super Tutor Pre Algebra (ESDPALG)

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

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

6x- 3 = y; (0, -3)

Here 6x - 3 = y is the equation

The ordered pair is (0, -3 )

It means if you substitutes 0 for x and -3 for y the equation will be true

Let us substitute and see:-

'6x' (means 6 times x) means 6 times 0 which is equal to 0

So the equation 6x - 3 = y becomes

0- 3 = -3 is correct

The ordered pair means the values of x and y which make the equation correct

Hope this is clear enough

Good luck

luciana44

Here 6x - 3 = y is the equation

The ordered pair is (0, -3 )

It means if you substitutes 0 for x and -3 for y the equation will be true

Let us substitute and see:-

'6x' (means 6 times x) means 6 times 0 which is equal to 0

So the equation 6x - 3 = y becomes

0- 3 = -3 is correct

The ordered pair means the values of x and y which make the equation correct

Hope this is clear enough

Good luck

luciana44

Sep 27, 2009 | The Learning Company Achieve! Math &...

2y - x = 3

x = 3y - 5

Add the two equations side by side,

2y - x + x = 3 + 3y - 5

2y = 3y - 2

y = 2

Plug this in the second equation to get x,

x = 3(2) - 5

x = 1

So the solution is**x = 1**, **y = 2**

or in ordered pair notation**(1, 2)**

x = 3y - 5

Add the two equations side by side,

2y - x + x = 3 + 3y - 5

2y = 3y - 2

y = 2

Plug this in the second equation to get x,

x = 3(2) - 5

x = 1

So the solution is

or in ordered pair notation

Jun 29, 2009 | Computers & Internet

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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