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Lady_Fox [76]
3 years ago
15

Why is the point-slope form useful if you want to write the equation of a line through a point that is parallel to a given line.

Mathematics
1 answer:
il63 [147K]3 years ago
6 0
Recall what a parallel line is; two lines that are parallel are defined as having the same gradient or slope. Consider a line:

y = mx + b

If we want to find a certain line that is / parallel / to the original line passing through an arbitrary point (x₁, y₁), it is useful to understand the point-gradient or point-slope formula.

The gradient to the line y = mx + b is simply m. So, any parallel line to y = mx + b will have the same gradient. Examples include: y = mx + 1, y = mx + 200, y = mx + g

All we need to know, now, is to identify what specific line hits the desired point. Well, the point-gradient formula can help with that. Recall that the point-gradient formula is:

y - y₀ = m(x - x₀), where (x₀, y₀) is the point of interest.
Hence, it is useful to use the point-slope formula when asked for a point and a set of parallel lines to the original line.
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a biologist counted 15 squirrels in 3 acres of forest. based on that data how many squirrels would be expected yo inhabit a 275
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Answer:


Step-by-step explanation:

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8 0
4 years ago
How do you do this problem? I need to know how you found the answer.
Alexxandr [17]

to get the equation of a line, we simply need two points, say for the Red one ... notice in the graph the lines passes through (0,2) and (-1,6), so let's use those


\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{-1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{-1-0}\implies \cfrac{4}{-1}\implies -4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=-4(x-0) \\\\\\ y-2=-4x\implies \blacktriangleright y=-4x+2 \blacktriangleleft


now, for the Blue one, say let's use hmmm it passes through (0,2) and (1.6)


\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-2}{1-0}\implies \cfrac{4}{1}\implies 4 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-2=4(x-0) \\\\\\ y-2=4x\implies \blacktriangleright y=4x+2 \blacktriangleleft

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Answer:

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Step-by-step explanation:

I divided them to make sure its right im not 100 percent sure tho

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