An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from #(2 ,9 )# to #(8 ,5 )# and the triangle's area is #48 #, what are the possible coordinates of the triangle's third corner?

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Answer 1 · 2018-06-30T14:53:06

#(161/13,235/13), or, (31/13,-53/13)#.

Let the third corner be at #(x,y)#.

Thus, the vertices of the #Delta# are, #(x,y), (2,9) and (8,5)#.

Hence, the area of the #Delta# is given by, #1/2|D|," where, "#

#D=|(x,y,1),(2,9,1),(8,5,1)|#,

#=x(9-5)-y(2-8)+1(10-72)#,

#=4x+6y-62#,

#=2(2x+3y-31)#.

But, this area is #48. :. |(2x+3y-31)|=48#.

#:. 2x+3y-31=+-48, i.e., #

# 2x+3y=79............(1), or, 2x+3y=-17............(2)#.

Also, the length of side #B=sqrt{(x-2)^2+(y-9)^2}#,

&, that of #C=sqrt{(x-8)^2+(y-5)^2}#.

Then, #B=C.................."[Given]"#,

#rArr x^2+y^2-4x-18y+85=x^2+y^2-16x-10y+89#.

# rArr 12x-8y=4, or, 3x-2y=1....................(3)#.

Solving #(1), &, (3)," gives, "(x,y)=(161/13,235/13)," whereas, "#

# (2), &, (3), (x,y)=(-31/13,-53/13)#.

Thus, the third corner is #(161/13,235/13), or, (31/13,-53/13)#.

Enjoy Maths.!