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Which equation can be used to graph the line that passes through (6,15) and is perpendicular to the graph y-7= 1/2 (x+2)

Which equation can be used to graph the line that passes through (6,15) and is perpendicular-example-1
User Jdunning
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1 Answer

7 votes

Answer:

Option A. y = –2x + 27

Explanation:

y – 7 = ½(x + 2)

We'll begin by calculating the slope of the equation above. This can be obtained as follow:

y – 7 = ½(x + 2)

Clear bracket

y – 7 = ½x + 1

Rearrange

y = ½x + 1 + 7

y = ½x + 8

Comparing

y = ½x + 8

with

y = mx + c

The slope (m) = ½

Next, we shall determine the slope (m2) of the equation perpendicular to the line. This can be obtained as follow:

For perpendicularity,

m1 × m2 = –1

m1 = ½

½ × m2 = –1

m2 /2 = –1

Cross multiply

m2 = 2 × –1

m2 = –2

Therefore, the slope of the equation perpendicular to the line is –2

Finally, we obtained the equation of the line as follow:

Coordinate = (6, 15)

x1 coordinate = 6

y1 coordinate = 15

Slope (m) = –2

y – y1 = m(x – x1)

y – 15 = –2(x – 6)

Clear bracket

y – 15 = –2x + 12

Rearrange

y = –2x + 12 + 15

y = –2x + 27

Therefore, the equation is y = –2x + 27

User Sumit Murari
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