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Posted on Jan 02, 2017

SOURCE: analytic geometry

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

Posted on Dec 15, 2008

SOURCE: I own a ti-84 plus silver edition, and when i

Hello,

And they are perpendicular. But due to the finite resolution of the screen they do not appear to be. However if you had a geometry application linked with the function graphing the measurement of the angle between the lines will show that it is 90 degrees, or 1.58 rad.

This screen capture from the TI'Nspire will convince you, I hope.

Hope it helps.

Posted on Nov 04, 2009

Testimonial: *"THanks for your help./ I was jsut trying to help my daughter and we werent sure why the lines didnt appear to be perpendicular."*

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.

- 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.

Apr 12, 2017 | Homework

Parallel lines are lines that go on forever and never cross. In mathematics terms, they have the same slope, but different y-intercepts. For example, y= 3x + 5 and y=3x +7 are parallel lines.

Perpendicular lines are lines that intersect each other at 90 degrees or at right angles. If y=5 and we are looking for a line that intersects it at 90 degrees (or right angles), the line would be x= 4.

Another example would be y = x. The perpendicular line would be y = -x.

To find the perpendicular line, the slope will be the negative reciprocal. For example, y = 3x + 5, the line perpendicular would be y = - 1/3 x + b.

Another way of looking at this is if you take the slopes of the two line and multiply them, you should get -1.

Let me know if you have any questions.

Good luck.

Paul

Perpendicular lines are lines that intersect each other at 90 degrees or at right angles. If y=5 and we are looking for a line that intersects it at 90 degrees (or right angles), the line would be x= 4.

Another example would be y = x. The perpendicular line would be y = -x.

To find the perpendicular line, the slope will be the negative reciprocal. For example, y = 3x + 5, the line perpendicular would be y = - 1/3 x + b.

Another way of looking at this is if you take the slopes of the two line and multiply them, you should get -1.

Let me know if you have any questions.

Good luck.

Paul

Jan 26, 2016 | Office Equipment & Supplies

2x + 7y - 14 = 0

One equation in 2 unknowns has many solutions. Generally you can just substitute x = 1,2,3,4,5 etc and find y. Let us stick to simple outcomes

7y = 14 - 2x

x = 7 y = 0

x = -7 y = 4

x = 14 y = -2

x = -14 y = 6

One equation in 2 unknowns has many solutions. Generally you can just substitute x = 1,2,3,4,5 etc and find y. Let us stick to simple outcomes

7y = 14 - 2x

x = 7 y = 0

x = -7 y = 4

x = 14 y = -2

x = -14 y = 6

Apr 11, 2014 | Mathsoft StudyWorks! Mathematics Deluxe...

3x-7y=2 is an equation of a line. That line doesn't go through the point (6, -7) though. Are you looking for the equation of a line through the point parallel to the first line? Perpendicular?

Jun 02, 2013 | Texas Instruments TI-84 Plus Calculator

Ok so first you need to find what x equals in terms of y. so use the first equation .2x+.7y=1.5 solve for x by the following steps

subract .7y from both sides: .2x= 1.5-.7y --->

divide by .2 on both sides: x= 7.5 - 3.5y. --->

Now plug in x= 7.5 - 3.5y for x in the second equation 0.3 (7.5 - 3.5y) -0.2y=1 --->

Distribute the .3: 2.25 - 1.05y -.2y = 1--->

solve for y by combining like terms (the y's) and subtractive 2.25: -1.25y = -1.25--->

divide by -1.25 and you get y= 1

Now go back to your x= 7.5 - 3.5y you solved for. Plug in 1 for y in the equation and solve for x--->

x=7.5-3.5(1) so x= 4 and y=1 yay your done!

subract .7y from both sides: .2x= 1.5-.7y --->

divide by .2 on both sides: x= 7.5 - 3.5y. --->

Now plug in x= 7.5 - 3.5y for x in the second equation 0.3 (7.5 - 3.5y) -0.2y=1 --->

Distribute the .3: 2.25 - 1.05y -.2y = 1--->

solve for y by combining like terms (the y's) and subtractive 2.25: -1.25y = -1.25--->

divide by -1.25 and you get y= 1

Now go back to your x= 7.5 - 3.5y you solved for. Plug in 1 for y in the equation and solve for x--->

x=7.5-3.5(1) so x= 4 and y=1 yay your done!

May 03, 2011 | Texas Instruments TI-30 XIIS Calculator

Hello,

And they are perpendicular. But due to the finite resolution of the screen they do not appear to be. However if you had a geometry application linked with the function graphing the measurement of the angle between the lines will show that it is 90 degrees, or 1.58 rad.

This screen capture from the TI'Nspire will convince you, I hope.

Hope it helps.

And they are perpendicular. But due to the finite resolution of the screen they do not appear to be. However if you had a geometry application linked with the function graphing the measurement of the angle between the lines will show that it is 90 degrees, or 1.58 rad.

This screen capture from the TI'Nspire will convince you, I hope.

Hope it helps.

Nov 04, 2009 | Texas Instruments TI-84 Plus Silver...

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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