Question

the value of k so that the line Containing the points (-8,k) and (-4,-8) is perpendicular) to the line containing the points (10, -11) and (5,-5)

Answer 1

Answer:

$k= \frac{-34}{3}$

Step-by-step explanation:

Step 1 :-

Perpendicular condition :-

Step 1:-

Two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

$m_{2} = \frac{-1}{m_{1} }$

$m_{1} m_{2} = -1$

Step 2:-

The given points are (-8,k) and (-4,-8)

$m_{1} = \frac{y_{2}-y_{1} }{x_{2}-x_{1} }$

finding slope of the first line is $m_{1}$

using formula $m_{1} = \frac{-8-k}{-4+8}$

=  $\frac{-8-k}{4}$

finding slope of the second line is $m_{2}$

using formula $m_{2} = \frac{--5+11}{5-10}$

=  $\frac{6}{-5}$

Step 3:-

using perpendicular condition

The two lines are perpendicular and their slopes are

$m_{1} m_{2} = -1$

$(\frac{-k-8}{4} )(\frac{-6}{5} )= -1$

simplification,we get solution is

$\frac{6(k+8)}{20} =-1$

$6 k+48 =-20$

$6 k = -20 -48$

$6 k = -68$

$\frac{-68}{6}$

$k= \frac{-34}{3}$

Final answer is

$k= \frac{-34}{3}$