Show that the points (12,8),(-2,6),&(6,0) are the vertices of y The right angled triangle . Also show that the mid point of the hypotenuse os equidistant from the angular points ?

Question

9326 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-22T12:02:40

As proved

Given vertices #A(12,8),B(-2,6),& C(6,0)#

#vec(AB) = sqrt ((12+2)^2 + (8-6)^2) = sqrt 200#

#vec(BC) = sqrt((-2-6)^2 + (6-0)^2) = 10#

#vec(CA) = sqrt((12-6)^2 + (6-0)^2) = 10#

#vec((AB)^2 )= 200 = vec((BC)^2) + vec((CA)^2 )= 100 + 100 = 200#

Hence it's a right triangle.

Also it's an isosceles triangle as #vec (BC) = vec(CA) = sqrt (100)#

Therefore, CD is a perpendicular bisector of hypotenuse AB and hence #vec(AD) = vec(BD)#

Answer 2 · 2018-02-22T12:12:23

#AB^2=BC^2+AC^2#, so it is right triangle.
# :. AD=BD=CD=sqrt50 :.D# is equidistant from angular
points.

Let the vertices of the triangle are #A(12,8),B(-2,6),C(6,0)#

Distance between two points #(x_1,y_1) and (x_2,y_2)# is

#D= sqrt((x_1-x_2)^2+(y_1-y_2)^2)#

#AB^2= (12+2)^2+(8-6)^2= 200# Similarly

#BC^2= (-2-6)^2+(6-0)^2= 100# and

#AC^2= (12-6)^2+(8-0)^2= 100 :. AB^2=BC^2+AC^2#

which are pythgorian tripples, property of right triangle of

which #AB# is hypotenuse. Let the mid point of hypotenuse

#AB# be #D# whose coordinate is #(12-2)/2, (8+6)/2#

#:. D# is #(5,7) :. AD=sqrt((12-5)^2+(8-7)^2)=sqrt50 #

#BD=sqrt((-2-5)^2+(8-7)^2)=sqrt50 #

#CD=sqrt((6-5)^2+(0-7)^2)=sqrt50#

# :. AD=BD=CD=sqrt50#. Hence #D# is equidistant from

angular points. [Ans]