Solving Pair of Equations

Question

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.

Answer 1

$2x-3y=6 \iff y=\color\red{\frac23}x-2$

$6x+ky=4 \iff y=\color\green{-\frac6k}x+\frac4k$

Perpendicular $\iff \left(\color\red{\frac23}\right)\cdot\left(\color\green{-\frac6k}\right)=-1 \iff k=4$

Answer 2

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 $k$.

Answer 3

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 $

Answer 4

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