Answer:
Let's see what to do buddy...
Step-by-step explanation:
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Step (1)
To find the distance between the line which formed like :

and the point which is like :

we use following equation :

_________________________________ Step (2)


So ;



°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°
A = ( 2 , - 1 )
°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°°
So we have :



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And we're done.
Thanks for watching buddy good luck.
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