Solving Pair of Equations This is an problem I came up on some website or another...
" Two perpendicular lines are represented by the equations $2x - 3y = 6$ and $6x + ky = 4$. What is the value of $k$? TOSS-UP"
Any help? $3$ variables, $2$ equations?? No idea how to solve.
I tried elimination by getting to $ky + 9y = -14$, and then $k(y+9) = - 14$, and then $-14/(y+9) = k$, but I don't know how to do this. 
 A: $2x-3y=6 \iff y=\color\red{\frac23}x-2$
$6x+ky=4 \iff y=\color\green{-\frac6k}x+\frac4k$
Perpendicular $\iff \left(\color\red{\frac23}\right)\cdot\left(\color\green{-\frac6k}\right)=-1 \iff k=4$
A: Two lines are perpendicular if their normals are perpendicular.
The normal of $2x - 3y = 6$ is $[2,-3]$, and the normal of $6x + ky = 4$ is $[6,k]$.
Set $[2,-3] \cdot [6,k] = 12 - 3k = 0$ and solve for $k$.
A: Slope of the first line : 
$$ \frac {2}{3} $$
Slope of the second line : 
$$ -\frac{6}{k} $$
Since they are perpendicular, the product of the slopes is $-1$. 
Hence :
$$
k = 4
$$
UPDATE :
The product of the slopes is -1 since :
If $\alpha $is the angle between the positive x axis and one of the given lines and $\beta $ is the angle between the positive x axis and the other line, for them to be perpendicular :
$$
|\beta - \alpha| = \frac{\pi}{2}
$$
Hence 
$$
-1 = tan(\alpha)tan(\beta)
$$
Or the product of the slopes should be $ -1 $
A: Equation of line y = mx + c
Where m is slope of line.

Line 1-

$2x-3y=6$
$3y=2x-6$
$y=\frac{2}{3}x-2$
So slope $m_{1}$ = $\frac{2}{3}$

Line 2-

$6x+ky=4$
$ky=-6x+4$
$y=\frac{-6}{k}x+\frac{4}{k}$
So slope $m_{2}$ = $\frac{-6}{k}$

Two lines are perpendicular so $m_{1} \times m_{2}$ = -1

$\frac{2}{3} \times \frac{-6}{k}$ = -1
k = 4
