Finding corresponding coordinate based on known ordered pair
$begingroup$
I need to know how to find the corresponding Y coordinate based on a given X coordinate that lies between two ordered pairs.
Example:
(60,25) and (70,40) Find the Y coordinate that corresponds to a X coordinate of 65
functions graph-theory
asked Jan 8 at 7:07
ArcheyArchey
1054
$endgroup$
add a comment |
$begingroup$
I need to know how to find the corresponding Y coordinate based on a given X coordinate that lies between two ordered pairs.
Example:
(60,25) and (70,40) Find the Y coordinate that corresponds to a X coordinate of 65
functions graph-theory
asked Jan 8 at 7:07
ArcheyArchey
1054
$endgroup$
add a comment |
$begingroup$
I need to know how to find the corresponding Y coordinate based on a given X coordinate that lies between two ordered pairs.
Example:
(60,25) and (70,40) Find the Y coordinate that corresponds to a X coordinate of 65
functions graph-theory
asked Jan 8 at 7:07
ArcheyArchey
1054
$endgroup$
I need to know how to find the corresponding Y coordinate based on a given X coordinate that lies between two ordered pairs.
Example:
(60,25) and (70,40) Find the Y coordinate that corresponds to a X coordinate of 65
functions graph-theory
functions graph-theory
asked Jan 8 at 7:07
ArcheyArchey
1054
asked Jan 8 at 7:07
ArcheyArchey
1054
asked Jan 8 at 7:07
ArcheyArchey
1054
asked Jan 8 at 7:07
ArcheyArchey
1054
asked Jan 8 at 7:07
ArcheyArchey
1054
1054
add a comment |
add a comment |
1 Answer
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$begingroup$
If the function is a straight line between two points $(x_1, y_1)$ and $(x_2, y_2)$, then the corresponding value for any $x$ is this equation
$y = y_1 + frac{x-x_1}{x_2-x_1}(y_2-y_1)$.
edited Jan 8 at 8:40
Archey
1054
answered Jan 8 at 7:42
Yanger MaYanger Ma
1014
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If the function is a straight line between two points $(x_1, y_1)$ and $(x_2, y_2)$, then the corresponding value for any $x$ is this equation
$y = y_1 + frac{x-x_1}{x_2-x_1}(y_2-y_1)$.
edited Jan 8 at 8:40
Archey
1054
answered Jan 8 at 7:42
Yanger MaYanger Ma
1014
$endgroup$
add a comment |
$begingroup$
If the function is a straight line between two points $(x_1, y_1)$ and $(x_2, y_2)$, then the corresponding value for any $x$ is this equation
$y = y_1 + frac{x-x_1}{x_2-x_1}(y_2-y_1)$.
edited Jan 8 at 8:40
Archey
1054
answered Jan 8 at 7:42
Yanger MaYanger Ma
1014
$endgroup$
add a comment |
$begingroup$
If the function is a straight line between two points $(x_1, y_1)$ and $(x_2, y_2)$, then the corresponding value for any $x$ is this equation
$y = y_1 + frac{x-x_1}{x_2-x_1}(y_2-y_1)$.
edited Jan 8 at 8:40
Archey
1054
answered Jan 8 at 7:42
Yanger MaYanger Ma
1014
$endgroup$
If the function is a straight line between two points $(x_1, y_1)$ and $(x_2, y_2)$, then the corresponding value for any $x$ is this equation
$y = y_1 + frac{x-x_1}{x_2-x_1}(y_2-y_1)$.
edited Jan 8 at 8:40
Archey
1054
answered Jan 8 at 7:42
Yanger MaYanger Ma
1014
edited Jan 8 at 8:40
Archey
1054
edited Jan 8 at 8:40
Archey
1054
edited Jan 8 at 8:40
Archey
1054
1054
answered Jan 8 at 7:42
Yanger MaYanger Ma
1014
answered Jan 8 at 7:42
Yanger MaYanger Ma
1014
answered Jan 8 at 7:42
Yanger MaYanger Ma
1014
1014
add a comment |
add a comment |
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