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777dan777 [17]
2 years ago
5

Find the slope of the line that passes through the points (2 -5) and (7, 1)

Mathematics
1 answer:
exis [7]2 years ago
5 0

Answer:

6/5

Step-by-step explanation:

We can find the slope given two points by

m = (y2-y1)/(x2-x1)

    = (1 - -5)/(7-2)

     = (1+5)/(7-2)

      =6/5

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

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