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Write an equation of a line that is perpendicular to y=-0.3+6 that passes through the point (3,-8)

1 Answer

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Answer:

y= 10/3 x -18

Explanation:

Perpendicular lines meet the following condition:

m2*m1=-1 (1)

The product of their slope is -1.

From the equation of the first line we can obtain the first slope.

y=-0.3x+6 -> m1=-0.3

Hence, the slope of our desired line is obtained from equation (1).

m2=10/3.

Remember, the equation of a line is given by:

y=mx+b

Now, we need to find the term 'b' and we will do that by evaluating the point (3,-8).

-8=10/3 * 3 +b.

b=-18.

Finally, our desired line has the following equation:

y= 10/3 x -18

User Huan Zhang
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