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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 \(2x + y = 13\)

 Mar 7, 2024

Best Answer 

 #1
avatar+14966 
+1

Find all points (x, y) that are 5 units away from the point (2, 7) and that lie on the line 2x + y = 13

Helo ABJelly!

 

\(f_1(x)=\pm\sqrt{5^2-(x-2)^2}+7\\ f_2(x)=-2x+13\\ f_1(x)=f_2(x)\\ \pm\sqrt{5^2-(x-2)^2}+7=-2x+13\\ \sqrt{5^2-(x-2)^2}\ ^2=(-2x+6)^2\\ 5^2-(x-2)^2=4x^2-24x+36\)

 

\(25-x^2+4x-4=4x^2-24x+36\\ 5x^2-28x+15=0\\ x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ \color{blue}x\in \{\frac{3}{5},5\}\\ \color{blue}f_1(x)=11.8\\ \color{blue}f_2(x)=3\)

 

The points we are looking for are (0.6,11.8) and (5,3).

 

laugh !

 Mar 7, 2024
 #1
avatar+14966 
+1
Best Answer

Find all points (x, y) that are 5 units away from the point (2, 7) and that lie on the line 2x + y = 13

Helo ABJelly!

 

\(f_1(x)=\pm\sqrt{5^2-(x-2)^2}+7\\ f_2(x)=-2x+13\\ f_1(x)=f_2(x)\\ \pm\sqrt{5^2-(x-2)^2}+7=-2x+13\\ \sqrt{5^2-(x-2)^2}\ ^2=(-2x+6)^2\\ 5^2-(x-2)^2=4x^2-24x+36\)

 

\(25-x^2+4x-4=4x^2-24x+36\\ 5x^2-28x+15=0\\ x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ \color{blue}x\in \{\frac{3}{5},5\}\\ \color{blue}f_1(x)=11.8\\ \color{blue}f_2(x)=3\)

 

The points we are looking for are (0.6,11.8) and (5,3).

 

laugh !

asinus Mar 7, 2024

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