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Show that the points (12,8) ,(-2,6) and (6,0) are the vertices of the right angled triangle and also show that the mid- points of the hypotenuse is equidistant from the angular points and Some students arranged a picnic the budget for food was rupees 240 to confusion for students of the group failed to go the cost of food for each student got increased by rupees 5 how many students went for the picnic

Show that the points (12,8) ,(-2,6) and (6,0) are the vertices of the right angled triangle and also show that the mid- points of the hypotenuse is equidistant from the angular points

 
and
 

Some students arranged a picnic the budget for food was rupees 240 to confusion for students of the group failed to go the cost of food for each student got increased by rupees 5 how many students went for the picnic

 

Grade:10

2 Answers

Harsh Patodia IIT Roorkee
askIITians Faculty 907 Points
8 years ago
A=(12,8)
B=(-2,6)
C=(6,0)
Using Distance formula
AB= sqrt(200)
BC= sqrt(100)
AC= sqrt(!00)

Clearly BC2 + AC2 = AB2

AB will be hypotenuse and similarly using distance formula you can prove the other thing.
CEL
11 Points
7 years ago
A(12,8) x1,y1
B(-2,6) x2,y2
C(6,0) x3,y3
We need to show that AB2=BC2+AC2
AB2= (x2-x1)2+(y2-y1)2
AB= sqrt(200)
 
similarly find the distance of BC and AC with the distance formula
BC2= (6+2)2+(0-6)2
BC= 10
AC2= (12-6)2+(8-0)2
AC= 10
 so triangle ABC is a right angled triangle through pythagoras theorem. 
 
Midpoint Q of hypotenuse AB = 12-2/2 , 8+6/2
=(5,7) x4,y4
We need to show that AQ=QB=QC
QC2=(x4-x3)2+(y4-y3)2
QC2=(5-6)2+(7-0)2
QC= sqrt(50)
AQ=QB
AQ=AB/2
=sqrt(200)/2
=2sqrt(50)/2
=sqrt(50)
 therefore AQ=BQ=QC
hence proved :)

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