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#?

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#?

Mason Adams · Accepted Answer

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

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]}

Claire Akin · Answer

undefined The equation of the pair of lines passing through the point (2,3) and perpendicular to the given line pair 2x^2 - xy - 6y^2 + 4x + 6y = 0 is:

$$ 2x^2 - xy - 6y^2 + 4x + 6y = 0 $$

To find the equation of the perpendicular lines passing through (2,3), we need to find the slope of the given line and then find the negative reciprocal of that slope to get the slope of the perpendicular lines.

The given line can be rewritten in the form $ y = mx + c $, where $ m $ is the slope:

$$ 2x^2 - xy - 6y^2 + 4x + 6y = 0 $$
$$ \Rightarrow 6y^2 - xy + 2x^2 + 6y - 4x = 0 $$
$$ \Rightarrow 6y^2 - xy + 2x^2 + 6y - 4x = 0 $$
$$ \Rightarrow 6y^2 - xy + 6y + 2x^2 - 4x = 0 $$
$$ \Rightarrow 6y(y + 1) - x(y + 1) + 2x(x - 2) = 0 $$
$$ \Rightarrow (6y - x)(y + 1) + 2x(x - 2) = 0 $$

Comparing with $ y = mx + c $, we have:

$$ m = \frac{6y - x}{y + 1} $$

At the point (2,3):

$$ m = \frac{6(3) - 2}{3 + 1} = \frac{18 - 2}{4} = \frac{16}{4} = 4 $$

Now, the negative reciprocal of the slope $ m $ is $ -\frac{1}{4} $, which is the slope of the perpendicular lines.

Using the point-slope form of the line equation:

$$ y - y_1 = m(x - x_1) $$
$$ y - 3 = -\frac{1}{4}(x - 2) $$
$$ y - 3 = -\frac{1}{4}x + \frac{1}{2} $$
$$ y = -\frac{1}{4}x + \frac{7}{2} $$

Therefore, the equation of the pair of lines passing through the point (2,3) and perpendicular to the given line is $ y = -\frac{1}{4}x + \frac{7}{2} $.