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Grade: 12th pass
        
Sir question number 14
........................................ Please 
2 months ago

Answers : (1)

Aditya Gupta
1690 Points
							
first draw the diagram. then you will find AB, BC and CA using distance formula. in fact AB= root 10, BC= AC= root 5.
so largest side is AB. now you ll find eqns of DE, EF and DF, and finally find their intersections to determine vertices D, E and F.
i ll sho u how 2 find eqn of DE and the remaining eqns of EF and DF u can find similarly.
eqn of AB= y – 2= 3(x – 7) or y= 3x – 19
let eqn of DE be y= 3x + c as it is parallel to AB.
now distance b/w parallel lines= |c1 – c2|/sqrt(1+m^2)
but given dist b/w DE and AB= sqrt 10
so sqrt 10= | – 19 – c|/sqrt(1+3^2)
or ± 10= c+19
or c= – 9 or – 29.
since DEF is external to ABC, with the help of figure u can see that DE has to be to the left of AB. now by the origin test you cn easily see that c= – 9 and – 29 is rejected.
so eqn of DE: y= 3x – 9
similarly proceed ahead.
kindly approve :)
2 months ago
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