An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from #(7 ,1 )# to #(8 ,5 )# and the triangle's area is #32 #, what are the possible coordinates of the triangle's third corner?

Answer 1

The possible points are #(-257/34, 115/17)# and #(767/34, -13/17)#

Given:

#"Area" = 32"units"^2#

The endpoints of side A are #(7,1)# and #(8,5)#.

Let side #A =# the base of the triangle.

Find the length of the base:

#"Base" = sqrt((8-7)^2+(5-1)^2)#

#"Base" = sqrt(1^2+4^2)#

#"Base" = sqrt17#

The area of a triangle is:

#"Area"=(1/2)("Base")("Height")#

Substitute in the values of Area and Base:

#32=(1/2)(sqrt17)("Height")#

#"Height" = 64sqrt17/17#

Compute midpoint of side A

#x_"mid"= (x_"end"+x_"start")/2 = (8+7)/2 = 15/2#

#y_"mid"= (y_"end"+y_"start")/2 = (5+1)/2 = 3#

Find the two points that are the same distance as height away from the midpoint:

#64sqrt17/17=sqrt((x-15/2)^2+(y-3)^2)" [1]"#

We need the slope of side A:

#m_a = (5-1)/(8-7)#

#m_a = 4#

The slope of the height is perpendicular to side A:

#m_h = -1/m_a#

#m_h = -1/4#

Use the point slope form of the equation of a line to write the equation for the height:

#y = -1/4(x-15/2)+3" [2]"#

With equations [1] and [2] this ugly, I am going to use WolframAlpha to solve them:

The possible points are #(-257/34, 115/17)# and #(767/34, -13/17)#

Here is an image of all of the parts of this problem that should show you that the answer is correct:

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Answer 2

#(-257/34,115/17), or, (767/34,-13/17)#

We solve this Problem deploying the methods of Analytic Geometry.

Supoose that the third vertex of the triangle is #P(x,y).#

Recall that, the Area of a Triangle having vertices

#(x_1,y_1), (x_2,y_2), and, (x_3,y_3)# is #1/2|D|,# where,
#D=|(x_1,y_1,1),(x_2,y_2,1),(x_3,y_3,1)|.#
Hence, in our case, since the area is #32,# we have, for
#32=1/2|D|, where, D=|(x,y,1),(7,1,1),(8,5,1)|#
#=x(1-5)-y(7-8)+1(35-8)=-4x+y+27.#
#:. 1/2*|-4x+y+27|=32 rArr -4x+y+27=+-64.#
#:.-4x+y=64-27=37 (ast^1) or -4x+y=-91(ast^2).# Also, Length of side B = Length of side C
#rArr (x-7)^2+(y-1)^2=(x-8)^2+(y-5)^2#
# rArr -14x+49-2y+1=-16x+64-10y+25#
#rArr2x+8y=39 rArr 4x+16y=78................(star).#
Solving #(ast^1) & (star), y=115/17, x=-257/34,# while,
#(ast^2) & (star) rArr y=-13/17, x=767/34.#
Thus, the possible third vertex can be #(-257/34,115/17), or, #
#(767/34,-13/17)#

Enjoy Maths.!

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Answer 3

Since the triangle is isosceles, let's denote the third corner as ( D ). ( BD ) and ( CD ) are equal sides.

The length of side ( A ) can be found using the distance formula between two points.

[ A = \sqrt{(8 - 7)^2 + (5 - 1)^2} = \sqrt{1^2 + 4^2} = \sqrt{17} ]

The area of the triangle can be found using the formula:

[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} ]

Given that the area is 32 and the base is ( \sqrt{17} ), we can solve for the height:

[ 32 = \frac{1}{2} \times \sqrt{17} \times \text{height} ]

[ \text{height} = \frac{64}{\sqrt{17}} ]

Now, the coordinates of point ( D ) can be found by moving up or down from point ( (8, 5) ) by a distance equal to ( \frac{64}{\sqrt{17}} ), keeping in mind that the triangle is isosceles, and hence the points ( B ) and ( C ) lie on a line perpendicular to ( AD ).

So, the possible coordinates of the third corner ( D ) are ( (8, 5 + \frac{64}{\sqrt{17}}) ) and ( (8, 5 - \frac{64}{\sqrt{17}}) ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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