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Slav-nsk [51]
3 years ago
8

Find the value of k so that 48x-ky=11 and (k+2)x+16y=-19 are perpendicular lines.

Mathematics
1 answer:
Rufina [12.5K]3 years ago
5 0

Answer: k = -1 +/- √769

<u>Step-by-step explanation:</u>

48x - ky = 11

<u>-48x        </u>  <u> -48x</u>

         -ky = -48x + 11

         \frac{-ky}{-k} = \frac{-48x}{-k} + \frac{11}{-k}    

           y =\frac{48x}{k} - \frac{11}{k}

Slope: \frac{48}{k}

*************************************************************************

 (k + 2)x + 16y = -19

<u>- (k + 2)x          </u>   -<u>(k + 2)x </u>

                 16y = -(k + 2)x - 19

                  \frac{16y}{16} = -\frac{(k + 2)x}{16} - \frac{19}{16}

                  y = -\frac{(k + 2)x}{16} - \frac{19}{16}

Slope: -\frac{(k + 2)}{16}

**********************************************************************************

\frac{48}{k} and -\frac{(k + 2)}{16} are perpendicular so they have opposite signs and are reciprocals of each other.  When multiplied by its reciprocal, their product equals -1.

-\frac{(k + 2)}{16} *  \frac{k}{48} = -1

\frac{(k + 2)k}{16(48)} = 1

Cross multiply, then solve for the variable.

(k + 2)(k) = 16(48)

k² + 2k - 768 = 0

Use quadratic formula to solve:

k = -1 +/- √769



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