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Find all points $(x,y)$ that are $5$ units away from the point $(2,7)$ and that lie on the line $y = 5x - 28.$

 Jun 4, 2024
 #1
avatar+729 
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Let the point $(x, y)$ be $5$ units away from the point $(2, 7)$. This means that the distance between $(x, y)$ and $(2, 7)$ is $5$ units.

 

Using the distance formula, we have


\[\sqrt{(x - 2)^2 + (y - 7)^2} = 5.\]


Squaring both sides, we get


\[(x - 2)^2 + (y - 7)^2 = 25.\]

Since the point $(x, y)$ lies on the line $y = 5x - 28$, we substitute $y = 5x - 28$ into the equation above and solve for $x$:


\[(x - 2)^2 + (5x - 35)^2 = 25.\]


\[x^2 - 4x + 4 + 25x^2 - 350x + 1225 = 25.\]


\[26x^2 - 354x + 1205 = 0.\]

The solutions to this quadratic equation for $x$ are $x = 5$ and $x = \frac{23}{13}$. Plugging these values of $x$ back into the equation $y = 5x - 28$, we find that the corresponding $y$ values are $y = 2$ and $y = 1$, respectively.

Therefore, the points that are $5$ units away from $(2, 7)$ and lie on the line $y = 5x - 28$ are $(5, 2)$ and $\left(\frac{23}{13}, 1\right)$.

 Jun 4, 2024
 #2
avatar+129747 
+1

Just a slight mistake by eramsby

 

The quadratic should be

 

26x^2 - 354x + 1204 = 0

 

The solutions  are  x  = 7    and x = 86/13

 

So y =  5(7) - 28  =  7

 

And  y = 5(86/13)  - 28  = 66/13

 

The points are  (7,7)  and (86/13 , 66/13)

 

 

cool cool cool

 Jun 4, 2024

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