3 years ago

Which line has a slope of 5/8 and goes through the point (8, 2)

1 answer:

6 0

The slope $m$ and a point $(x_{0},\;y_{0})$ that the line goes through, give us a equation for the line:

$y-y_{0}=m\cdot (x-x_{0})\\\\ y-2=\dfrac{5}{8}\cdot (x-8)\\\\\\ y-2=\dfrac{5}{8}\,x-\dfrac{5}{8}\cdot 8\\\\\\ y-2=\dfrac{5}{8}\,x-5\\\\\\ y=\dfrac{5}{8}\,x-5+2\\\\\\ \boxed{\begin{array}{c} y=\dfrac{5}{8}\,x-3 \end{array}}$

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A line that contains the point (-8,-1) and is perpendicular to -15x +3y = 9

zaharov [31]

Put the other line in point slope form...(solve for y :D )

3y=15x+9

y=5x+3, so this line has a slope of 5

For a line to be perpendicular to another the slopes have to be negative reciprocals of each other...mathematically...

s1*s2=-1

So for the other line to be perpendicular to the first:

5m=-1, m=-1/5, thus our other line (so far) is:

y=b-x/5, now we can use the point (-8,-1) to solve for the y-intercept, "b":

-1=b-(-8)/5

-1=b+8/5

-1=b+1.6

b=-2.6

So the perpendicular line that contains the point (-8,-1) is:

y=-0.2x-2.6

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nalin [4]

C. 1 solution hope it helps

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Anarel [89]

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3 years ago