Question

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

Answer 1

C

Step-by-step explanation:

Let $l_{1}$ be the equation of the line with slope $m_{1}$ and let $l_{2}$ be the equation of the line with slope $m_{2}$.

As we know,for two lines $l_{1}$ and $l_{2}$ to be perpendicular,

$m_{1}\times m_{2} =-1$

In the given problem,slope of given line=$m_{1}$=$\frac{-6}{-1}=6$

For option a,slope =$m_{2}$$=\frac{-6}{1} =-6$

In this case,$m_{1}\times m_{2}=6\times -6=-36$

So,option a is incorrect.

For option b,slope =$m_{2}$$=\frac{6}{1} =6$

In this case,$m_{1}\times m_{2}=6\times 6=36$

So,option b is incorrect.

For option c,slope =$m_{2}$$=\frac{-1}{6} =\frac{-1}{6}$

In this case,$m_{1}\times m_{2}=6\times \frac{-1}{6}=-1$

So,option c is correct.

For option d,slope =$m_{2}$$=\frac{1}{6} =\frac{1}{6}$

In this case,$m_{1}\times m_{2}=6\times \frac{1}{6}=1$

So,option d is incorrect.

Answer 2

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 $\frac{1}{6} \times \frac{-1}{6} \neq -1$

b. x - 6y = -52   ⇒  -6y = -52  - x

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

but $\frac{1}{6} \times \frac{1}{6} \neq -1$

c. 6x + y = -52    

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

$\frac{1}{6} \times (-6) = -1$

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 $\frac{1}{6} \times 6 \neq -1$

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