What is the midpoint and distance between points P(3,5) and Q(7,5)

Mathematics

1 answer:

storchak [24]3 years ago

4 0

The formula of a midpoint:

$M\left(\dfrac{x_1+x_2}{2};\ \dfrac{y_1+y_2}{2}\right)$

The formula of a distance between two points:

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

We have the points P(3, 5) and Q(7, 5). Substitute:

$x=\dfrac{3+7}{2}=\dfrac{10}{2}=5\\\\y=\dfrac{5+5}{2}=\dfrac{10}{2}=5$

$d=\sqrt{(5-5)^2+(7-3)^2}=\sqrt{0^2+4^2}=\sqrt{4^2}=4$

<h3>Answer:</h3><h3>Midpoint = (5, 5)</h3><h3>Distance = 4</h3>

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<u>SOLUTION: </u>

Given that, we have to find what is the length side of a triangle that has vertices at (-5, -1), (-5, 5), and (3, -1)

We know that, distance between two points $P\left(x_{1}, y_{1}\right) \text { and } Q\left(x_{2}, y_{2}\right)$ is given by

$P Q=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Now,

$\begin{array}{l}{\text { Distance between }(-5,-1) \text { and }(-5,5)=\sqrt{(-5-(-5))^{2}+(5-(-1))^{2}}} \\\\ {\qquad \begin{array}{l}{=\sqrt{(-5-(-5))^{2}+(5-(-1))^{2}}} \\\\ {=\sqrt{(0)^{2}+(5+1)^{2}}=\sqrt{(6)^{2}}=6} \\\\ {=\sqrt{(-5)^{2}+(5+1)^{2}+(5-(-1))^{2}}} \\\\ {=\sqrt{(-8)^{2}+(5+1)^{2}}=\sqrt{64+36}=\sqrt{100}=10} \\\\ {=\sqrt{(3-1)^{2}+(-1-(-1))^{2}}} \\\\ {=\sqrt{(5+3)^{2}+(0)^{2}}=\sqrt{(8)^{2}}=8}\end{array}}\end{array}$