Answer:
As the product of the slop of both lines is -1.
- [tex]m_1\times m_2=-1[/tex]
- [tex]3\times \frac{-1}{3}=-1[/tex]
Therefore, the given equations are perpendicular.
Step-by-step explanation:
Given the equations
[tex]-3x+y=-4[/tex]
[tex]x+3y=6[/tex]
The slope-intercept form of the equation is
[tex]y=mx+b[/tex]
where m is the slope and b is the y-intercept.
Writing both equations in the slope-intercept form
[tex]-3x+y=-4[/tex]
[tex]y=3x-4[/tex]
So by comparing with the slope-intercept form we can observe that
slope of equation = 3
i.e.
[tex]m_1=3[/tex]
also
[tex]x + 3y = 6[/tex]
[tex]3y\:=\:6-x[/tex]
[tex]y=-\frac{1}{3}x+2[/tex]
So by comparing with the slope-intercept form we can observe that
the slope of equation = -1/3
i.e.
[tex]m_2=-\frac{1}{3}[/tex]
as
The slope of the perpendicular line is basically the negative reciprocal of the slope of the line.
so
The slope [tex]m_2[/tex] is the negative reciprocal of the slope [tex]\:m_1[/tex]
Also, the product of two perpendicular lines is -1.
i.e.
[tex]m_1\times m_2=-1[/tex]
VERIFICATION:
It is clear that the product of the slop of both lines is -1.
[tex]m_1\times m_2=-1[/tex]
[tex]3\times \frac{-1}{3}=-1[/tex]
Therefore, the given equations are perpendicular.
