Question

Answer the following:

Answer 1

Answer:

As the product of the slop of both lines is -1.

• $m_1\times m_2=-1$

• $3\times \frac{-1}{3}=-1$

Therefore, the given equations are perpendicular.

Step-by-step explanation:

Given the equations

$-3x+y=-4$

$x+3y=6$

The slope-intercept form of the equation is

$y=mx+b$

where m is the slope and b is the y-intercept.

Writing both equations in the slope-intercept form

$-3x+y=-4$

$y=3x-4$

So by comparing with the slope-intercept form we can observe that

slope of equation = 3

i.e.

$m_1=3$

also

$x + 3y = 6$

$3y\:=\:6-x$

$y=-\frac{1}{3}x+2$

So by comparing with the slope-intercept form we can observe that

the slope of equation = -1/3

i.e.

$m_2=-\frac{1}{3}$

as

The slope of the perpendicular line is basically the negative reciprocal of the slope of the line.

so

The slope $m_2$ is the negative reciprocal of the slope $\:m_1$

Also, the product of two perpendicular lines is -1.

i.e.

$m_1\times m_2=-1$

VERIFICATION:

It is clear that the product of the slop of both lines is -1.

$m_1\times m_2=-1$

$3\times \frac{-1}{3}=-1$

Therefore, the given equations are perpendicular.