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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 15, 2024

Best Answer 

 #1
avatar+1250 
+1

First, let's make some observations. 

Note that all points that are 5 units away from the point (2, 7) froms a circle with radius 5 and center (2, 7). 

The equation for that circle would be \((x-2)^2+(y-7)^2=25\)

 

Combining this with the second equation, we get a system. 

\((x-2)^2+(y-7)^2=25\\ y=5x-28\)

 

We already have a y value in terms of x, so we plug that into the first equation. 

We get

\((x-2)^2+(5x-35)^2=25\\ x^2-4x+4+25x^2-350x+1225=25\\ 13x^{2}-177x+602=0\)

 

Now, we can solve for x. Using the quadratic equation, we get

\(x=\frac{-{(-177})\pm \sqrt{{(-177)}^{2}-4\cdot {13}\cdot{602}}}{2\cdot {13}}\)

\(x=\frac{177\pm 5}{26}\)

\(x=7\\ x=\frac{86}{13}\)

 

Plugging these values for y, we get 

\(\begin{pmatrix}x=7,\:&y=7\\ x=\frac{86}{13},\:&y=\frac{66}{13}\end{pmatrix}\)

 

So our final answer is (7, 7) and (86/13, 66/13)

 

Thanks! :)

 Jun 15, 2024
 #1
avatar+1250 
+1
Best Answer

First, let's make some observations. 

Note that all points that are 5 units away from the point (2, 7) froms a circle with radius 5 and center (2, 7). 

The equation for that circle would be \((x-2)^2+(y-7)^2=25\)

 

Combining this with the second equation, we get a system. 

\((x-2)^2+(y-7)^2=25\\ y=5x-28\)

 

We already have a y value in terms of x, so we plug that into the first equation. 

We get

\((x-2)^2+(5x-35)^2=25\\ x^2-4x+4+25x^2-350x+1225=25\\ 13x^{2}-177x+602=0\)

 

Now, we can solve for x. Using the quadratic equation, we get

\(x=\frac{-{(-177})\pm \sqrt{{(-177)}^{2}-4\cdot {13}\cdot{602}}}{2\cdot {13}}\)

\(x=\frac{177\pm 5}{26}\)

\(x=7\\ x=\frac{86}{13}\)

 

Plugging these values for y, we get 

\(\begin{pmatrix}x=7,\:&y=7\\ x=\frac{86}{13},\:&y=\frac{66}{13}\end{pmatrix}\)

 

So our final answer is (7, 7) and (86/13, 66/13)

 

Thanks! :)

NotThatSmart Jun 15, 2024

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