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Suppose that there is only one value of \(x\) for which the distance from \((5,6)\) to \((3x-7,ax+2)\) is \(4\). If a \(a \neq 0,\) what is \(a\)?

 Nov 27, 2020
 #1
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Suppose that there is only one value of x  for which the distance from \(P_1(5,6) \)  to \(P_2(3x-7,ax+2) \)  is 4 . If  \(a \neq 0, \) what is  a?


 

Hello Guest!

 

\(d^2=(y_2-y_1)^2+(x_2-x_1)^2\)

\(4^2=(ax+2-6)^2+(3x-7-5)^2\\ 16=(ax-4)^2+(3x-12)^2\\ 16=a^2x^2-8ax+16+9x^2-72x+144\\ f(x)=(a^2+9)x^2-(8a+72)x+144=0\)

 

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(x = {(8a+72) \pm \sqrt{{\color{blue} (8a+72)^2-4\cdot (a^2+9)\cdot 144}} \over 2\cdot (a^2+9)}\)    

  

When the radical is zero, the function f (x) has a single solution.

 

\( (8a+72)^2-4\cdot (a^2+9)\cdot 144=0 \)      | \(a \neq 0 \)

\(a=2.25\)

\(x = 3.2\)

laugh  !

 Nov 28, 2020
edited by asinus  Nov 28, 2020
edited by asinus  Nov 28, 2020
edited by asinus  Nov 28, 2020
 #2
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Thank you!

Guest Nov 28, 2020

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