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3 years ago

Find the midpoint of the line segment defined by the points: (5, 4) and (−2, 1)

Mathematics

2 answers:

Serhud [2]3 years ago

8 0

Answer: The mid point of the line segment is (1.5,2.5).

Step-by-step explanation:

Since we have given that

A (5,4) and B(-2,1) are the end points of the line segment.

We need to find the mid point say 'C':

As we know the formula for "Mid point ":

$C=(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2})\\\\C=\dfrac{5-2}{2},\dfrac{4+1}{2})\\\\C=(\dfrac{3}{2},\dfrac{5}{2})\\\\C=(1.5,2.5)$

Hence, the mid point of the line segment is (1.5,2.5).

anyanavicka [17]3 years ago

3 0

$\bf ~~~~~~~~~~~~\textit{middle point of 2 points }\\ \quad \\
\begin{array}{ccccccccc}
&&x_1&&y_1&&x_2&&y_2\\
% (a,b)
&&(~{{ 5}} &,&{{ 4}}~) 
% (c,d)
&&(~{{ -2}} &,&{{ 1}}~)
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left( \cfrac{-2+5}{2}~~,~~\cfrac{1+4}{2} \right)\implies \left(\frac{3}{2}~~,~~\frac{5}{2} \right)\implies \left(1\frac{1}{2}~~,~~2\frac{1}{2} \right)$

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Leno4ka [110]

1 step: $S_{5}=T_{1}+T_{2}+T_{3}+T_{4}+T_{5}$ , $S_{3}=T_{1}+T_{2}+T_{3}$ , then
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2 step: $T_{n}=T_{1}*q^{n-1}$ , then
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3 step: Solve this system $\left \{ {{T_1*q^{3}*(q^{2}-1)= \frac{5}{2} } \atop {T_{1}*q^{3}*(q+1)=5} \right.$ and dividing first equation on second we obtain $\frac{q^{2}-1}{q+1}= \frac{ \frac{5}{2} }{5}$ . So, $\frac{(q-1)(q+1)}{q+1} = \frac{1}{2}$ and $q-1= \frac{1}{2}$ , $q= \frac{3}{2}$ - the common ratio.

4 step: Insert $q= \frac{3}{2}$ into equation $T_{1}*q^{3}*(q+1)=5$ and obtain $T_{1}* \frac{27}{8}*( \frac{3}{2}+1 ) =5$ , from where $T_{1}= \frac{16}{27}$ .

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3 years ago

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Readme [11.4K]

Answer:

∠1 = 112°

Step-by-step explanation:

∠2 = ∠3 = 112° ( corresponding angles )

∠1 = ∠2 = 112° ( vertically opposite angles )

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