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slava [35]
3 years ago
13

If the ordinate is three more than twice the abscissa and x {-1, 0, -1/3}, which of the following sets of ordered pairs satisfie

s the given conditions?
A. {(-1, 1), (0, 3), (-1/3, 7/3)}
B. {(-1, 1), (0, 3), (-1/3, 11/3)}
C. {(-1, -1), (0, 3), (-1/3, 1)}
Mathematics
2 answers:
Molodets [167]3 years ago
8 0

Answer:

c

Step-by-step explanation:

lions [1.4K]3 years ago
7 0

Answer:

I think it is A not sure though

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xxMikexx [17]

\text{Let}\ k:y=m_1x+b_1\ \text{and}\ l:y=m_2x+b_2\\\\l\perp k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\\text{We have the points J(-24, -4) and K(-4, 6)}.\\\\\text{The formula of a slope:}\\\\m=\dfrac{y_2-y_1}{x_2-x_1}\\\\\text{substitute:}\\\\m_1=\dfrac{6-(-4)}{-4-(-24)}=\dfrac{10}{20}=\dfrac{1}{2}\\\\\text{therefore}\ m_2=-\dfrac{1}{\frac{1}{2}}=-2\\\\\text{The formula of a midpoint:}\\\\\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)\\\\\text{substitute the coordinates of the points J and K:}

x=\dfrac{-24+(-4)}{2}=\dfrac{-28}{2}=-14\\\\y=\dfrac{-4+6}{2}=\dfrac{2}{2}=1\\\\\text{midpoint}\ (-14,\ 1)\\\\\text{The point-slope form:}\\\\y-y_1=m(x-x_1)\\\\\text{substitute}\ m=-2,\ x_1=-14\ \text{and}\ y_1=1:\\\\y-1=-2(x-(-14))\\\\\boxed{y-1=-2(x+14)}

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