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SVEN [57.7K]
2 years ago
15

Write an equation in standard form for the line that passes through the given points.

Mathematics
1 answer:
Agata [3.3K]2 years ago
8 0

So, pretend this is your x-axis and y-axis:

                                   I

        I

   (-2,7) •  I

        I

        I       • (2, 5)

        I

        I

        I  

        I

 _________________I____________________

        I

        I

        I

TO GET FROM POINT (-2, 7) TO POINT (2, 5), WE MOVE DOWN 2 AND OVER 4, SO THE SLOPE IS -1/2. IF WE FOLLOW THAT SLOPE AND MOVE DOWN 1 AND OVER 2 FROM THE FIRST POINT OF  (-2, 7), WE WILL LAND ON A POINT LOCATED AT (0, 6), WHICH WOULD BE THE "Y-INTERCEPT". WE WERE JUST ABLE TO CALCULATE THE SLOPE OF THE LINE AND THEN USE THE SLOPE TO FIND THE INTERCEPT. SO, THE "SLOPE-INTERCEPT" FORM OF THE EQUATION FOR THIS LINE IS:

y = -1/2x + 6

TO RE-WRITE THIS IN STANDARD FORM, WE JUST WANT TO MOVE THE X VARIABLE OVER TO THE LEFT WITH THE Y VARIABLE, SO:

y = -1/2x + 6

+1/2x   + 1/2x

1/2x + y = 6 .... and that is your answer!

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2 years ago
​ Quadrilateral ABCD ​ is inscribed in this circle.
Anna [14]
A quadrilateral, has 4 sides and its internal angles sum, add up to 360, now... you have 3 angles give.. .but we don't have C

so.. C is the difference of all the three angles from 360 or  \bf \measuredangle C=360-x-(2x+1)-148\implies \measuredangle C=360-x-2x-1-148   whatever that is, now, you'll get some value in x-terms

so.... now once we know what C is

you can if you want, do a search in google for "inscribed quadrilateral conjecture",  I can do a quick proof if you need one

but in short, for a quadrilateral inscribed in a circle, each pair of opposites angles are "supplementary angles", namely they add up to 180°

so.. what the dickens does all that mean?   well D+B=180 and A+C = 180

now. we know what A is, 2x+1
and by now, you'd know what C is from  360-x-2x-1-148

so... add them together then   and

\bf \begin{array}{cccclllll}
A&+&C&=&180\\
\uparrow &&\uparrow \\
(2x+1)&+&(360-x-2x-1-148)&=&180
\end{array}

solve for "x"


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