Question

Can someone check my answer?

Answer 1

I think it's PQ & Plane PQS. Am I correct?

Answer 2

sorry, what is the question, more exactly?

Answer 3

Nameing the line and plane shown in the picture. @Michele_Laino

Answer 4

Yeah you're correct

Answer 5

yes! you are right!

Answer 6

Ok thank you!

Answer 7

:)

Answer 8

Could you help me with a problem?

Answer 9

@Michele_Laino

Answer 10

yes!

Answer 11

@Michele_Laino

Answer 12

the coordinates of the midpoint of a segment whose ends are:
\((x_1,y_1)\), and \((x_2,y_2)\), is:
\[\Large M = \left( {\frac{{{x_1} + {x_2}}}{2},\frac{{{y_1} + {y_2}}}{2}} \right)\]

Answer 13

oops.. are*

Answer 14

so it's (2,2) right?

Answer 15

yes!

Answer 16

can you help with one last problem?

Answer 17

ok!

Answer 18

here we have to compute these three distances:
\(d(A,B),\;d(B,C),\;d(CA)\)

Answer 19

for example, here is the distance \(d(A,B)\):
\[\Large d\left( {A,B} \right) = \sqrt {{{\left( { - 3 - 9} \right)}^2} + {{\left( {7 - \left( { - 2} \right)} \right)}^2}} = ...?\]

Answer 20

Can you help more? lol I don't get it.

Answer 21

hint:
\[\Large \begin{gathered}
d\left( {A,B} \right) = \sqrt {{{\left( { - 3 - 9} \right)}^2} + {{\left( {7 - \left( { - 2} \right)} \right)}^2}} = \hfill \\
\hfill \\
= \sqrt {{{12}^2} + {9^2}} = \sqrt {144 + 81} = ...? \hfill \\
\end{gathered} \]

Answer 22

It's 21 units right?

Answer 23

it is:
\[\Large \begin{gathered}
d\left( {A,B} \right) = \sqrt {{{\left( { - 3 - 9} \right)}^2} + {{\left( {7 - \left( { - 2} \right)} \right)}^2}} = \hfill \\
\hfill \\
= \sqrt {{{12}^2} + {9^2}} = \sqrt {144 + 81} = \sqrt {225} = 15 \hfill \\
\end{gathered} \]

Answer 24

next:
\[\Large \begin{gathered}
d\left( {B,C} \right) = \sqrt {{{\left( { - 3 - \left( { - 3} \right)} \right)}^2} + {{\left( { - 2 - 7} \right)}^2}} = \hfill \\
\hfill \\
= \sqrt {{0^2} + {9^2}} = ...? \hfill \\
\hfill \\
\hfill \\
d\left( {C,A} \right) = \sqrt {{{\left( { - 3 - 9} \right)}^2} + {{\left( { - 2 - \left( { - 2} \right)} \right)}^2}} = \hfill \\
\hfill \\
= \sqrt {{{12}^2} + {0^2}} = ...? \hfill \\
\end{gathered} \]