Premium Online Home Tutors
3 Tutor System
Starting just at 265/hour
Answer :
The green flag is \(\frac{1}{4} th \) of total distance =
\(\frac{1}{4} × 100 = 25 \) m in second line
\(\therefore \) the coordinate of green flag
= (2 , 25 )
Similarly coordinate of red flag
= (8 , 20)
Distance between red and green flag
= \( \sqrt{(8-2)^2 + (20-25)^2} \)
\( = \sqrt{6^2+(-5)^2} \)
\( = \sqrt{36+25} \)
\( = \sqrt{61} m \)
Now blue flag is posted at the mid point of two flag
Then let, the coordinate of blue flag = (x , y )
\(\therefore( x , y ) =( \frac{(8+2)}{2} , \frac{20+25}{2} ) \)
\( =( 5 , \frac{45}{2} )= ( 5, 22.5 ) \)
\(\therefore \) the blue flag is posted in fifth line at a distance of 22.5 m