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julia-pushkina [17]
3 years ago
12

What is the slope-intercept form of a line that passes through points (2, 11) and (4, 17)?

Mathematics
2 answers:
butalik [34]3 years ago
5 0

Answer:

y = 3x + 5

Step-by-step explanation:

First, find the <em>rate of</em><em> </em><em>change</em><em> </em>[<em>slope</em>]:

-y₁ + y₂\-x₁ + x₂ = m

\frac{-11 + 17}{-2 + 4} = \frac{6}{2} = 3

Now, plug the coordinates into the Slope-Intercept Formula instead of the <em>Point-Slope</em><em> </em><em>Formula</em><em> </em>because you get it done much swiftly. It does not matter which ordered pair you choose:

17 = 3[4] + b

12

5 = b

y = 3x + 5

__________________________________________________________

11 = 3[2] + b

6

5 = b

y = 3x + 5

You see? I told you it did not matter which ordered pair you choose because you will always get the exact same result.

I am joyous to assist you anytime.

navik [9.2K]3 years ago
4 0

\bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{11})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{17}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{17-11}{4-2}\implies \cfrac{6}{2}\implies 3 \\\\\\ \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-11=3(x-2)\implies y-11=3x-6

\bf y=3x+5\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}

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Ivahew [28]
Let's go through the choices one by one

------------------------------------------
Choice A

If all sides are congruent, then this figure is a rhombus (by definition). If all angles are congruent, then we have a rectangle. Combine the properties of a rhombus with the properties of a rectangle and we have a square.

In terms of "algebra", you can think
rhombus+rectangle = square

Or you can draw out a venn diagram. One circle represents the set of all rhombuses; another circle represents the set of all rectangles. The overlapping region is the set of all squares. The overlapping region is inside both circles at the same time.

So we can rule out choice A. This guarantees we have a square when we want something that isn't a guarantee.

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Choice B

If we had a parallelogram with perpendicular diagonals, then we can prove that we have a rhombus (all four sides congruent). However, we don't know anything about the four angles of this parallelogram. Are they congruent? We don't know. So we can't prove this figure is a rectangle. The best we can say is that it's a rhombus. It may or may not be a rectangle. There isn't enough info about the rectangle & square part.

This is why choice B is the answer. We have some info, but not enough to be guaranteed everytime.

------------------------------------------
Choice C

This is a repeat of choice A. Having "all right angles" is the same as saying "all angles congruent". This is because "right angle" is the same as saying "90 degrees". So we can rule out choice C for identical reasons as we did with choice A.

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Choice D

As mentioned before in choice A, if we know that a quadrilateral is a rectangle and a rhombus at the same time, then the figure is also a square. This is always true, so we are guaranteed to have a square. We can cross choice D off the list.

------------------------------------------

Once again, the final answer is choice B


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