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# Algebra

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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

#1
+1280
+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
+1280
+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! :)

NotThatSmart Jun 15, 2024