Question about Texas Instruments TI-84 Plus Calculator

Ad

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

Posted on Oct 20, 2010

Ad

Hi there,

Save hours of searching online or wasting money on unnecessary repairs by talking to a 6YA Expert who can help you resolve this issue over the phone in a minute or two.

Best thing about this new service is that you are never placed on hold and get to talk to real repairmen in the US.

Here's a link to this great service

Good luck!

Posted on Jan 02, 2017

Ad

The "slope intercept form of the equation of a (the) line" is y=mx+b, where m is the slope of the line and b is the y-intercept.

We are given the slope of 1/2, so m= 1/2.

We can now write y=1/2 x + b.

Since the point (-2,-3) is on the line, we can substitute it in and solve for b. We put the -2 in for x and -3 in for y.

-3 = 1/2(-2) +b

-3 = -1 + b

-3 + 1 = -1 + b +1

-2 =b

Thus, the equation of the line is y= 1/2 x -2

To check if we did this correctly, plug in the point (-2, -3) to see if it works.

Left Side Right Side

-3 = 1/2 (-2) -2

= -1-2

= -3

We are given the slope of 1/2, so m= 1/2.

We can now write y=1/2 x + b.

Since the point (-2,-3) is on the line, we can substitute it in and solve for b. We put the -2 in for x and -3 in for y.

-3 = 1/2(-2) +b

-3 = -1 + b

-3 + 1 = -1 + b +1

-2 =b

Thus, the equation of the line is y= 1/2 x -2

To check if we did this correctly, plug in the point (-2, -3) to see if it works.

Left Side Right Side

-3 = 1/2 (-2) -2

= -1-2

= -3

Feb 24, 2015 | Office Equipment & Supplies

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...

**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...

You are looking for a line (y=m*x+b) and have two points. From this information you can generate two equations with two unknowns (m and b are unknown).

First plug in the first point (1,5) to the general form:

5(the y value of the point) = m*1(the x-value at this point)+b

Do the same for the second point you're given.

From here solve the first equation for m in terms of b.

Plug this value of 'm' into the second equation so you will end up with something like:

3=(something in terms of b)*(-2)+b

This final equation can be solved for b (try factoring)

You now have a value for the y-intercept. Plug that into y=m*x+b

Choose either of the two points, plug into the equation on the last line with the value of b known

You then know y, x, and b and have m as the remaining 1 unknown. Solve for that and put it all together for your final answer.

First plug in the first point (1,5) to the general form:

5(the y value of the point) = m*1(the x-value at this point)+b

Do the same for the second point you're given.

From here solve the first equation for m in terms of b.

Plug this value of 'm' into the second equation so you will end up with something like:

3=(something in terms of b)*(-2)+b

This final equation can be solved for b (try factoring)

You now have a value for the y-intercept. Plug that into y=m*x+b

Choose either of the two points, plug into the equation on the last line with the value of b known

You then know y, x, and b and have m as the remaining 1 unknown. Solve for that and put it all together for your final answer.

Apr 23, 2011 | Texas Instruments TI-83 Plus Calculator

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

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

276 people viewed this question

Usually answered in minutes!

×