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.