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trapecia [35]
3 years ago
13

If the point (4,-2) is included in a direct variation relationship, which point also belongs in this direct variation? A. (-4,2)

B. (-4,-2) C. (2,-4) D. (-2,4)
Mathematics
1 answer:
IceJOKER [234]3 years ago
5 0

Answer:

(-4,2) belongs in this direct variation.

Step-by-step explanation:

Let the direct variation relationship is expressed by the equation y = mx ........ (1), where x and y are in direct variation and k is the variation constant.

Now, the point (4,-2) is included in the direct variation relationship, then from equation (1) we get, -2 = 4m

⇒ m =  - \frac{1}{2}

Therefore, the equation (1) becomes y = - \frac{1}{2}x

⇒ x + 2y = 0 ......... (2)

Now, from the given four options only the point (-4,2) satisfies the relation (2).

Hence, (-4,2) belongs in this direct variation. (Answer)

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3 years ago
The midpoint of RS¯¯¯¯¯¯¯ is M. Use the given endpoint R(23, 14) and midpoint M(13, 8) to find the coordinates of the other endp
Oksanka [162]

The coordinates of other endpoint S is (3, 2)

<h3><u>Solution:</u></h3>

Given that midpoint of RS is M

Given endpoint R(23, 14) and midpoint M(13, 8)

To find: coordinates of the other endpoint S

<em><u>The formula for midpoint is given as:</u></em>

For a line containing containing two points (x_1, y_1) and (x_2, y_2) midpoint is given as:

m(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)

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m(x, y) = (13, 8)

(x_1, y_1) = (23, 14)\\\\(x_2, y_2) = ?

Substituting the given values in above formula, we get

(13,8)=\left(\frac{23+x_{2}}{2}, \frac{14+y_{2}}{2}\right)

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\begin{aligned}&13=\frac{23+x_{2}}{2} \text { and } 8=\frac{14+y_{2}}{2}\\\\&26=23+x_{2} \text { and } 16=14+y_{2}\\\\&x_{2}=3 \text { and } y_{2}=2\end{aligned}

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3 years ago
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