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# Linear Algebra Math Problem

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Question:

There is only one value of k for which the line x = k intersects the graphs of  y = x^2 + 6x + 5 and y = mx + b at two points which are exactly 5 units apart. If the line y = mx + b passes through the point (1, 6), and b ≠ 0, find the equation of the line. Enter your answer in the form "y = mx + b".

What I tried:

Since x = k, I substituted x for k into the equation y = x^2 + 6x + 5. Therefore, y = k^2 + 6k + 5, and y = mk + b. When this system of equations is solved, the two points will be 5 units apart. Since I know that the coordinates (1, 6) pass through y = mx + b, I know that 6 = m + b.

I can also factor y = k^2 + 6k + 5 as y = (k + 1)(k + 5). This also means that (k + 1)(k + 5) = mx + b.

This is all I have figured out...

Aug 21, 2022

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There is only one value of k for which the line x = k intersects the graphs of  y = x^2 + 6x + 5 and y = mx + b at two points which are exactly 5 units apart. If the line y = mx + b passes through the point (1, 6), and b ≠ 0, find the equation of the line. Enter your answer in the form "y = mx + b".

For convenience  let    the  vertical  line be  x  = r

Then the point on the parabola  that this line intersects =  (r , r^2 + 6r + 5)

Call the point on the parabola   that the line of interest intersects  =  (q , q^2 + 6q + 5)

The slope of this line =   (6 - [ q^2 + 6q + 5 ])  / ( 1 - q)  =    ( 1 - q^2 - 6q) / (1 -q)  =  (q^2 + 6q - 1) / ( q - 1)  =  m

You  have correctly figured out the  6 = m + b

So 6 - m = b

So  6 -  (q^2 + 6q - 1) / (q - 1)   = b

So  [ 6 (q -1) -  (q^2 + 6q - 1) ]  / (q -1)  = b

So  [ q^2 + 5  ] / (1 - q)  =  b

Then   the point   (0, b)  =    [ 0 , q^2 + 5 ] / [1 -q)]   is on  the line

The slope between   this  point and (1,6)  =      [  6 - q^2 - 5 ] / [  (1-q)  ]  =  (1 - q^2] / (1 - q)  =  1 + q

Equating slopes

( 1 + q)  = ( q^2 + 6q -1) / (q -1)

q^2 - 1  = q^2 + 6q - 1

0 = 6q        →   q =  0

So   the line of interest intersects the parabola at    [ 1-q , q^2 + 5  ]  =  ( 0,5)

So  m  = slope between (0, 5) and (6, 1)  =    (6 -5) / (1 -0) =  1

And

b = 5

Our line  is   y = x + 5    { verify that   (1,6)   is on this  line }

When  x = r

Then y =  r + 5

And we need the distance  between   (r , r+5  )    and  ( r,  r^2 + 6r + 5)   to equal   5

So

sqrt  [   (r - r)^2 + [ (r +5) - (r^2 + 6r + 5) ]^2  ] =  5

A little tricky to solve....but..solving this  for the  r  we need produces

r ≈ .8541

So the vertical line is  x =  .8541

And y  ≈

(.8541)^2  + 6(.8541) + 5  ≈ 10.8541

So   ( .8541, 10.8541)  =  "B"  on the  graph  = the place where the vertical line crosses the parabola

And

(r , r + 5)  =   ( .85410 , 5.8541) = "B"  on the  graph  where the vertical line intersects y = x + 5

Here's a picture :    Aug 22, 2022