finding the equation from graph..
Keu ki bolte parba jodi kono somikoroner lekhor koekta bindu deoa thake ,tahole ar theke somikoron ta bair kora jabe ki...jemon(2,3),(5,8) ar somikoron ki hobe...
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- Joined:Fri Feb 18, 2011 11:30 pm
Re: finding the equation from graph..
$(y-3)/(x-2)=(y-8)/(x-5)=Slope(constant)$
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- Posts:78
- Joined:Thu Jan 20, 2011 10:46 am
Re: finding the equation from graph..
The equation through two points \[(x_{1},y_{1}) and (x_{2},y_{2})\]
is \[\frac{x-x_{1}}{x_{1}-x_{2}}=\frac{y-y_{1}}{y_{1}-y_{2}}\]
then rest is yours two points are given now find the equation
is \[\frac{x-x_{1}}{x_{1}-x_{2}}=\frac{y-y_{1}}{y_{1}-y_{2}}\]
then rest is yours two points are given now find the equation
Re: finding the equation from graph..
You have to be given enough points to be sure about the graph. From only two points, you can't be sure about the equation. For example, $(2,3)$ and $(5,8)$ lie on the line $5x-3y-1=0$ also on the circle with center $(6,4)$ and radius $\sqrt{17}$ and on infinitely many equations.rafid wrote:Keu ki bolte parba jodi kono somikoroner lekhor koekta bindu deoa thake ,tahole ar theke somikoron ta bair kora jabe ki...jemon(2,3),(5,8) ar somikoron ki hobe...
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Re: finding the equation from graph..
For example , if you are given three points ,you can always draw a circle(sometimes with center at point of infinity ) and given five points , you can always draw a conic.Zzzz wrote: You have to be given enough points to be sure about the graph. From only two points, you can't be sure about the equation. For example, $(2,3)$ and $(5,8)$ lie on the line $5x-3y-1=0$ also on the circle with center $(6,4)$ and radius $\sqrt{17}$ and on infinitely many equations.
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