Which equation has a graph that is perpendicular to the graph of -x + 6y = -12?

a. x + 6y = -67
b. x - 6y = -52
c. 6x + y = -52
d. 6x - y = 52

Respuesta :

Answer:

c) 6x + y = -52  is required equation perpendicular to the given equation.

Step-by-step explanation:

If the equation is of the form    : y = mx  + C.

Here m = slope of the equation.

Two equations are said to be perpendicular if the product of their respective slopes is -1.

Here, equation 1 :  -x + 6y = -12

or, 6y = -12  + x

or, y = (x/6)  - 2

Slope of line 1 = (1/6)

Now, for equation 2  to be  perpendicular:

Check for each equation:

a. x + 6y = -67       ⇒  6y = -67  - x

or, y = (-x/6)  - (67/6)      ⇒Slope of line 2 = (-1/6)

but [tex]\frac{1}{6} \times \frac{-1}{6}  \neq -1[/tex]

b. x - 6y = -52   ⇒  -6y = -52  - x

or, y = (x/6)  + (52/6)      ⇒Slope of line 2 = (1/6)

but [tex]\frac{1}{6} \times \frac{1}{6}  \neq -1[/tex]

c. 6x + y = -52    

or, y =y = -52  - 6x      ⇒Slope of line 2 = (-6)

[tex]\frac{1}{6} \times (-6)  =  -1[/tex]

Hence, 6x + y = -52  is required equation 2.

d. 6x - y = 52  ⇒  -y = 52  - 6x

or, y = 6x   - 52      ⇒Slope of line 2 = (6)

but [tex]\frac{1}{6} \times 6  \neq -1[/tex]

Hence,  6x + y = -52  is  the  only required equation .

C

Step-by-step explanation:

Let [tex]l_{1}[/tex] be the equation of the line with slope [tex]m_{1}[/tex] and let [tex]l_{2}[/tex] be the equation of the line with slope [tex]m_{2}[/tex].

As we know,for two lines [tex]l_{1}[/tex] and [tex]l_{2}[/tex] to be perpendicular,

[tex]m_{1}\times m_{2} =-1[/tex]

In the given problem,slope of given line=[tex]m_{1}[/tex]=[tex]\frac{-6}{-1}=6[/tex]

For option a,slope =[tex]m_{2}[/tex][tex]=\frac{-6}{1} =-6[/tex]

In this case,[tex]m_{1}\times m_{2}=6\times -6=-36[/tex]

So,option a is incorrect.

For option b,slope =[tex]m_{2}[/tex][tex]=\frac{6}{1} =6[/tex]

In this case,[tex]m_{1}\times m_{2}=6\times 6=36[/tex]

So,option b is incorrect.

For option c,slope =[tex]m_{2}[/tex][tex]=\frac{-1}{6} =\frac{-1}{6}[/tex]

In this case,[tex]m_{1}\times m_{2}=6\times \frac{-1}{6}=-1[/tex]

So,option c is correct.

For option d,slope =[tex]m_{2}[/tex][tex]=\frac{1}{6} =\frac{1}{6}[/tex]

In this case,[tex]m_{1}\times m_{2}=6\times \frac{1}{6}=1[/tex]

So,option d is incorrect.

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