# Find the equation of the pair of lines passing through the point (2,3) and perpendicular to the line pair 2x^2-xy-6y^2+4x+6y=0?

Jan 3, 2018

$6 {x}^{2} - x y - 2 {y}^{2} - 21 x + 14 y = 0$

#### Explanation:

Given: $2 {x}^{2} - x y - 6 {y}^{2} + 4 x + 6 y = 0$

Factor:

$\left(2 x + 3 y\right) \left(x - 2 y + 2\right) = 0$

The equation of each line is:

$2 x + 3 y = 0$ and $x - 2 y = - 2$

To make lines that are perpendicular, we swap the coefficients and change the sign of one coefficient:

$3 x - 2 y = {C}_{1}$ and $2 x + y = {C}_{2}$

To find the values of ${C}_{1}$ and ${C}_{2}$ substitute the point $\left(2 , 3\right)$ into each equation:

$3 \left(2\right) - 2 \left(3\right) = {C}_{1}$ and $2 \left(2\right) + \left(3\right) = {C}_{2}$

${C}_{1} = 0$ and ${C}_{2} = 7$

The equation of the lines are:

$3 x - 2 y = 0$ and $2 x + y = 7$

Write second line so that it is equal to 0:

$3 x - 2 y = 0$ and $2 x + y - 7 = 0$

Multiply the two lines:

$\left(3 x - 2 y\right) \left(2 x + y - 7\right) = 0$

$3 x \left(2 x + y - 7\right) - 2 y \left(2 x + y - 7\right) = 0$

$6 {x}^{2} + 3 x y - 21 x - 4 x y - 2 {y}^{2} + 14 y = 0$

$6 {x}^{2} - x y - 2 {y}^{2} - 21 x + 14 y = 0$

graph{(2x^2-xy-6y^2+4x+6y)(6x^2-xy-2y^2-21x+14y)=0 [-10.25, 9.75, -3.64, 6.36]}