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Sidana [21]
1 year ago
12

Consider the following points on cartesian plane. A (1;2) BC3; 1) and CG-3; 4) prove that Points A, B and C are collinear. ​

Mathematics
1 answer:
Arlecino [84]1 year ago
5 0

Step-by-step explanation:

Collinear points have the same gradient

find the gradient(slope) between AB first

= \frac{1 - 2}{3 - 1}  \\  =  \frac{ - 1}{2}

Find the one for BC

= \frac{4 - 1}{ - 3 - 3}  \\  =  \frac{3}{ - 6}  \\  =  \frac{ - 1}{2}

Find the gradient between AC now

=  \frac{4 - 2}{ - 3 - 1}  \\  =  \frac{2}{ - 4}  \\  =  \frac{ - 1}{2}

YOUR LAST STATEMENT NOW WILL BE

POINTS A,B AND C ARE COLLINEAR SINCE THE GRADIENTS BETWEEN THEM IS THE SAME!!!

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HELP ASAP IM STRUGGLING : The instructions on a box of chicken patties state that one patty should be cooked for 2.5 minutes in
AveGali [126]

Answer:

y= 1.5x + 2.5

Step-by-step explanation:

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4 0
3 years ago
Write the equation for a line that has an initial value of 3 and 3/4 as it’s rate of change
frutty [35]

the slope goes by several names

• average rate of change

• rate of change

• deltaY over deltaX

• Δy over Δx

• rise over run

• gradient

• constant of proportionality

however, is the same cat wearing different costumes.

initial value of 3, namely when x = 0, y = 3, so we have the point (0 , 3) and it has a rate or slope of 3/4.

(\stackrel{x_1}{0}~,~\stackrel{y_1}{3})\qquad \qquad \stackrel{slope}{m}\implies \cfrac{3}{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-\stackrel{y_1}{3}=\stackrel{m}{\cfrac{3}{4}}(x-\stackrel{x_1}{0})\implies y=\cfrac{3}{4}x+3

3 0
3 years ago
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