a perimeter of a triangle is 30 units and the points (0, -5) and (0,5) are two of its vertices. Find the graph?

Question

a perimeter of a triangle is 30 units and the points (0, -5) and (0,5) are two of its vertices. Find the graph of the third vertex.

Answer 1

10 + sqrt((x^2 + (y - 5)^2) + sqrt(x^2 + (y + 5)^2)) = 30



sqrt(x^2 + (y - 5)^2 + x^2 + (y + 5)^2) = 20



sqrt(2x^2 + 2y^2 + 50) = 20



2x^2 + 2y^2 + 50 = 400



x^2 + y^2 = 350



This is a circle with radius sqrt(350) and center (0,0)



.

Answer 2

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the given points lie in the y-axis and equidistant from the origin (0, 0).

the distance between the two points is 10,

so a point in the x-axis (either on the + or - side) that is 10 units from either point can be the third vertex.

the x & y - axes and the line from one of the given points to this point in the x-axis form a right triangle.

using the pythagorean theorem which states that

....."the square of the hypotenuse is equal to the sum of the squares of its sides.", we have

5 ² + x ² = 10 ²

x = √ 75

...= ± 8.66



now, any point in the plane with the sum of its distance from the two given points equal to 20 can also be the third vertex (except of course the two points that lie on the y-axis which are +10 & -10).



if you have a string that is 20 units long with the ends fixed at the given points and a pen stretching the string and moved around the plane you will have an ellipse that crosses the x-axis at +8.66 & -8.66 and the y-axis at +10 & -10. this ellipse would contain all the possible third vertex of the triangle.



the equation of the ellipse whose major and minor axes coincide with the Cartesian axes is

( x ² / a ² ) + ( y ² / b ² ) = 1

...................where a = x-intercept = 8.66

...............................b = y -ntercept = 10

( x ² / 8.66 ² ) + ( y ² / 10 ² ) = 1

( x ² / 75 ) + ( y ² / 100 ) = 1

........................multiply both sides of the equation by 300;

4x ² + 3y ² = 300

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☻☻☻☻☻☻☻☻☻☻

edit for some comments:



Dr G's method of solving the graph of the third vertex is also right but with a minor error:



" 30=sqrt[x^2+(y+5)^2]+sqrt[x^2+(5-y)^2] "



the 30 should be 20, because the equation does not include the distance between the two given points which is 5+5=10.





let's simplify this:



√ [ x ² + ( y + 5 ) ² ] + √ [ x ² + ( 5 - y ) ² ] = 20

√ [ x ² + ( y + 5 ) ² ] = 20 - √ [ x ² + ( 5 - y ) ² ]

.....................................square both sides of the equation;

x ² + ( y + 5 ) ² = 400 - 40√ [ x ² + ( 5 - y ) ² ] + x ² + ( 5 - y ) ²

x ² + y ² + 10y + 25 = 400 - 40√ [ x ² + ( 5 - y ) ² ] + x ² + y ² - 10y + 25

.....................................x ² + y ² + 25 would cancel out.

.....................................add 10y to, & subtract 400 from, both sides of the equation;

20y - 400 = 40√ [ x ² + ( 5 - y ) ² ]

.....................................divide both sides of the equation by 20;

y - 20 = 2√ [ x ² + ( 5 - y ) ² ]

.....................................square both sides of the equation;

y ² - 40y + 400 = 4 [ x ² + ( 5 - y ) ² ]

y ² - 40y + 400 = 4 ( x ² + y ² - 10y + 25 )

y ² - 40y + 400 = 4x ² + 4y ² - 40y + 100

.....................................- 40y would cancel out.

.....................................subtract y ² & 100 from both sides of the equation;

300 = 4x ² + 3y ²

.....or

4x ² + 3y ² = 300

░░░░░░░░░░..........☺☺☺☺☺..........b i n g o ! ! !

Answer 3

the third point could be anywhere such that its the sum of its distance from point a (0.-5) and point b (0,5), is 20 (since the distance from a to b is already 10).



the distance to point a from any given point, (x,y), will be sqrt[x^2+(y+5)^2] (found by the pythagorean theorem; this might make more sense if you sketch the two points)



likewise, the distance to b will be sqrt[x^2+(5-y)^2]



the graph of all points where the perimeter will be 30 is the graph 30=sqrt[x^2+(y+5)^2]+sqrt[x^2+(5-y)^2]

Answer 4

Try a physical simulation.

Stick a couple of pins in a piece of graph paper. Take a piece of string 30 units long and loop it around the pins. Stick a pencil in the loop and trace the extents it can go.



You should get a specific type of conic section.

Answer 5

The perimeter = √8 + √20 + √42 = ....wait..