Question

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`?

Answer 1

`6x^2 - x y - 2 y^2 - 21 x + 14 y = 0`

Explanation:

Given: `2x^2-xy-6y^2+4x+6y=0`

Factor:

`(2x+3y)(x-2y+2) = 0`

The equation of each line is:

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

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

`3x-2y = C_1` and `2x+y=C_2`

To find the values of `C_1` and `C_2` substitute the point `(2,3)` into each equation:

`3(2)-2(3) = C_1` and `2(2)+(3)=C_2`

`C_1=0` and `C_2 = 7`

The equation of the lines are:

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

Write second line so that it is equal to 0:

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

Multiply the two lines:

`(3x-2y)(2x+y-7)=0`

`3x(2x+y-7)-2y(2x+y-7)=0`

`6x^2+3xy-21x-4xy-2y^2+14y=0`

`6x^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]}