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can one of you unhelpful users answer the question without putting links as a d a m n answer?
1. A line goes through the points (1, 6) and (–2, 8). Write the equation of the line in Point-Slope form. Show your work algebraically for full credit.











2. A line goes through the points (1, 6) and (–2, 8). Write the equation of the line in Slope-intercept form. Show your work algebraically for full credit. Hint: this is the same line as in #1.

Sagot :

Answer:

1) [tex]y-6=-\frac{2}{3} (x-1)[/tex] and 2) [tex]y=-\frac{2}{3} x+\frac{20}{3}[/tex]

Step-by-step explanation:

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Point-slope form is written as: [tex]y-y_1=m(x-x_1)[/tex]

Slope intercept form is written as: [tex]y=mx+b[/tex]

We also need to find the slope before putting it in this form, which can be done by [tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]

1) Points given are (1,6) and (-2,8)

Slope: [tex]m=\frac{8-6}{-2-1} =-\frac{2}{3}[/tex]

You can choose either of the two points to put into the formula. I just chose (1,6)

Answer: [tex]y-6=-\frac{2}{3} (x-1)[/tex]

2) Same points as before and we have the same slope as before. All we need is to find the y intercept which is our b in the equation. You can easily do this by plugging in a point and solving for b. I will choose point (1,6) again.

[tex]6=-\frac{2}{3} (1)+b[/tex]

[tex]6=-\frac{2}{3} +b[/tex]

[tex]6+\frac{2}{3}=-\frac{2}{3} +\frac{2}{3} +b[/tex]

[tex]b=6+\frac{2}{3}=\frac{20}{3}[/tex]

Answer: [tex]y=-\frac{2}{3} x+\frac{20}{3}[/tex]

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