What is the equation of the line passing through $(93, 78)$ $(93,78)$ and $(- 68, 44)$ $(-68,44)$ ?

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What is the equation of the line passing through $(93, 78)$ $(93,78)$ and $(- 68, 44)$ $(-68,44)$ ?

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Answer 1 · 2015-11-14T00:22:36

Find the line in the form $y = m x + b$ $y = mx + b$ .
The slope can be found through the formula $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ $m=(y_2-y_1)/(x_2-x_1)$ .
Thus, $m = \frac{44 - 78}{- 68 - 93} = \frac{- 34}{- 161} = \frac{34}{161}$ $color(red)(m)=(44-78)/(-68-93)=(-34)/-161=color(red)(34/161)$

Now, find $b$ $b$ by plugging $m$ $m$ into $y = m x + b$ $y = mx + b$ with one of the points.

With the point $(93, 78)$ $(93,78)$ : $78 = (\frac{34}{161}) 93 + b$ $78 = (34/161)93+b$
Multiply: $78 = \frac{3162}{161} + b$ $78 = 3162/161+b$
Find a common denominator: $\frac{12558}{161} = \frac{3162}{161} + b$ $12558/161=3162/161+b$
Subtract $\frac{3162}{161}$ $3162/161$ from both sides: $\frac{9396}{161} = b$ $color(red)(9396/161=b)$
This cannot be simplified.

Plug back into $y = m x + b$ $y=mx+b$ : $y = \frac{34}{161} x + \frac{9396}{161}$ $color(red)(y=34/(161)x+9396/161)$

This can also be written as $y = \frac{34 x + 9396}{161}$ $y=(34x+9396)/161$

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What is the equation of the line passing through $(93, 78)$ $(93,78)$ and $(- 68, 44)$ $(-68,44)$ ?

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What is the equation of the line passing through (93,78)(93,78) and (−68,44)(-68,44)?

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What is the equation of the line passing through $(93, 78)$ $(93,78)$ and $(- 68, 44)$ $(-68,44)$ ?