... triangle. Use evidence to support your claim. If it is not an equilateral triangle, what changes can be made to make it equilateral? Be specific
Determine if triangle DEF with coordinates D (3, 2), E (4, 6), and F (7, 3) is an equilateral?
- Posted:
- 3+ months ago by christynr...
- Topics:
- geometry, support, triangle, evidence
Details:
Answers (1)
Triangle DEF with coordinates
D (3, 2), E (4, 6), F (7, 3)
and line segments
d = EF, e = DF, f = DE
Question 1:
Is triangle DEF an equilateral triangle?
No, it isn't.
Use evidence to support your claim!
Pythagoras tells us that, if the line segment p = P1P2 with P1(x1, y1), P2(x2, y2), then
p² = (x1 - x2)² + (y1 - y2)². Hence
d² = (4 - 7)² + (6 - 3)² = 18 ->
d = sqrt(18). Likewise
e = sqrt(17)
f = sqrt(17).
So, d > e, d > f. Therefore, triangle DEF is not equilateral.
Question 2:
What changes can be made to make it equilateral?
Triangle DEF is not equilateral, but it is an isosceles triangle with
base d and legs e = f. Thus, the perpendicular bisector of d is also the bisectrix of the angle in D. It passes through the intersection point of every isosceles triangle with base d, including the one of the two equilateral triangles we are looking for.
If the line segment p = P1P2 with P1(x1, y1), P2(x2, y2), then the slope m of p is:
m_p = (y2 - y1) / (x2 -x1)
The slope m of d is, therefore:
m_d = -1
The slope of the perpendicular bisector d' of d,
equals the negative reciprocal of m_d:
m_d' = - (m_d)^-1 = 1.
Now we have a point and a slope and can build the linear equation of d':
d' := y - 2 = x - 3
y = x - 1 # !!!!! # Soon, this equation will become important.
Imagine you had constructed the line with E, F and its perpendicular bisector with compass and straightedge. In order to make equilateral triangles you would then adjust your compass to draw a circle with radius r = d. The intersections of this circle with the perpendicular bisector would then give you exactly the points you need.
We can do this algebraically.
The circle c with centre point C(xC, yC) and radius r has the equation:
c := (x - xC)² + (y - yC)² = r².
In our case C = E and r = d (you can take C = F as well, the result is the same).
(x - 4)² + (y - 6)² = 18
And now we substitute y with x-1 from the linear equation above.
(x - 4)² + (x - 1 - 6)² = 18
(x - 4)² + (x - 7)² = 18
x² - 8x + 16 + x² - 14x + 49 = 18
2x² - 22x + 47 = 0
x² - 11x + (47 / 2) = 0 ;;; and solve the quadratic equation
x1,2 = 11/2 +- sqrt (121/4 - 94/4)
x1,2 = 11/2 +- sqrt (27/4)
x1,2 = 11/2 +- sqrt ( (9/4) * 3)
x1,2 = 11/2 +- (3/2) sqrt (3)
x1 = 5.5 + 3/2 sqrt(3) = 8.09807...
x2 = 5.5 - 3/2 sqrt(3) = 2.90192... , that's the one you are looking for.
Insert x2 in the linear equation y=x-1
Your "true" D ( 5.5 - 3/2 sqrt(3), 4.5 - 3/2 sqrt(3) ).
(Btw: Here you can easily see that the coordinates of midpoint of d, M_d(5.5, 4.5) and, therefore,
3/2 sqrt(3) = sqrt (18) * sin (60) * sin (45)
3/2 sqrt(3) = sqrt (3*3*2) * sqrt (3) /2 * 1 / sqrt (2)
3/2 sqrt(3) = 3 * sqrt (2) / sqrt (2) * sqrt (3) / 2
3/2 sqrt(3) = 3 * sqrt (3) / 2 = 3/2 sqrt (3)
is the length of the cathetes of the slope triangle DM_d
M_d is the symmetrical center of the two equilateral triangles).
We have d with length sqrt(18) and slope m_d = -1. And we want to algebraically construct an equilateral triangle. Now, an equilateral triangle is also an equiangular triangle, each inner angle in the triangle being 180° / 3 = 60°. We can, therefore, construct the equations of the two missing sides e, f, set them equal and thereby determine point D.
As m_d = -1, the corresponding angle is tan^-1 (-1) = -45°.
e will then be at an angle (rotate d counter-clockwise in point F)
angle (e) = -45° + 60° = 15°.
And f will be at an angle (rotate d counter-clockwise in point E)
angle (f) = -45° + 120° = 75°.
To determine the corresponding slopes, we have to find
m_e = tan(15) = tan(45 - 30) , and
m_f = tan(75) = tan(45 + 30).
Because of
tan(a) = sin(a) / cos(a) and
tan(a + b) = (tan(a) + tan(b)) / (1 - tan(a) * tan(b)), and in analogy
tan(a - b) = (tan(a) - tan(b)) / (1 + tan(a) * tan(b)) we get
tan(45) = sin(45) / cos(45) = (1 / sqrt(2)) / (1 / sqrt(2)) = 1
tan (30) = sin(30) / cos(30) = (1 / 2) / (sqrt(3) / 2) = 1 / sqrt(3) and thus
m_e = tan(15) = 2 - sqrt(3)
m_f = tan(75) = 2 + sqrt(3)
[As this is here a side issue, I didn't type the steps. But if you need help with the derivations of these tangents, feel free to ask.]
With m_e and F(7,3) we can now define
e := y - 3 = (2 - sqrt(3)) (x - 7)
y = 2x - sqrt(3) * x - 11 + 7 * sqrt(3)
With m_f and E(4,6) we can also define
f := y - 6 = (2 + sqrt(3)) (x - 4)
y = 2x + sqrt(3) * x - 2 - 4 * sqrt(3).
Setting e = f yields
2x - sqrt(3) * x - 11 + 7 * sqrt(3) = 2x + sqrt(3) * x - 2 - 4 * sqrt(3)
-9 + 11 sqrt(3) = 2x sqrt(3)
(-9 + 11 sqrt(3)) / sqrt(3) = 2x
(-9 sqrt(3) + 33) / 3 = 2x
-3/2 sqrt(3) + 5.5 = x_D ;; x-coordinate of point D
Insert x_D into, say, the equation of e to get y_D
y = (-3/2 sqrt(3) + 5.5) (2 - sqrt(3)) - 11 + 7 sqrt(3)
= (-3 - 5.5 + 7) sqrt(3) + 4.5
y_D = -3/2 sqrt(3) + 4.5
D(-3/2 sqrt(3) + 5.5, -3/2 sqrt(3) + 4.5).
You could also find true D as one of the intersection points of two circles with centre points E, F and radius r = sqrt(18):
(x - 4)² + (y - 6)² = 18 ;; 1.a
(x - 7)² + (y - 3)² = 18 ;; 1.b
x² - 8x + 16 + y² - 12y + 36 = 18 ;; 2.a
x² - 14x + 49 + y² - 6y + 9 = 18 ;; 2.b
x² - 8x + y² - 12y + 52 = 18 ;; 3.a
x² - 14x + y² - 6y + 58 = 18 ;; 3.b , subtract 3.b from 3.a
6x - 6y - 6 = 0 ;; 4.
y = x - 1 ;; 5. , and lo and behold, there is an old acquaintance
And now you proceed by substituting y with x-1 in equation 1.a or 1.b same way as shown above ...