BC is a triangle right angled at A with vertices A,B,C in the anti clockwise sense in the order. A(1,2).B(-3,1) and vertex C lies on the X axis, BCEF is a square with vertices B,C,E,F in the clockwise sense in that order. ACD is an equilateral triangle with vertices A,C,D in the anti clockwise sense in that order. slope of AF is
Answers
Step-by-step explanation:
The square doesn't make sense. I'll assume they meant:
BCEF is a square with vertices B,C,E,F in the clockwise sense.
Besides, there's no way that D can be a vertex of the square with side BC and a vertex of the equilateral triangle with side AC.
A = (1,2)
B = (−3,1)
C = (x,0)
Slope AB = (2−1)/(1+3) = 1/4
Since ABC is right angled at A, then line AB is perpendicular to line AC
Slope AC = −1/(1/4)
−2/(x−1) = −4
1/(x−1) = 2
x − 1 = 1/2
x = 3/2
C = (3/2, 0)
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Since BCEF is a square with vertices B,C,E,F in the clockwise order:
E is point B rotated 270° clockwise (90° counter clockwise) about C
F is point C rotated 270° clockwise (90° counter clockwise) about E
(x,y) rotated 90° counter clockwise about point (a,b)
(x,y) ---> (a+b−y, b−a+x)
E = B(−3,1) rotated 90° ccw about C(3/2, 0)
E = (3/2+0−1, 0−3/2−3) = (1/2, −9/2)
F = C(3/2, 0) rotated 90° ccw about E(1/2, −9/2)
F = (1/2−9/2−0, −9/2−1/2+3/2) = (−4, −7/2)
E = (1/2, −9/2)
F = (−4, −7/2)
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A = (1, 2)
C = (3/2, 0)
D = (x, y)
Since ACD is an equilateral triangle then
AD = CD = AC
AD² = CD² = AC²
(x−1)² + (y−2)² = (x−3/2)² + (y−0)² = (3/2−1)² + (0−2)²
x² − 2x + 1 + y² − 4y + 4 = x² − 3x + 9/4 + y² = 1/4 + 4
x² − 2x + y² − 4y + 5 = x² − 3x + 9/4 + y² = 17/4
x² − 2x + y² − 4y + 5 = x² − 3x + 9/4 + y²
−2x − 4y + 5 = −3x + 9/4
x = 4y − 11/4
x² − 3x + 9/4 + y² = 17/4
(4y − 11/4)² − 3(4y − 11/4) + 9/4 + y² = 17/4
16y² − 22y + 121/16 − 12y + 33/4 + y² − 2 = 0
17y² − 34y + 221/16 = 0
1/16 (16y² − 32y + 13) = 0
y = 1 ± √3/4 = (4±√3)/4
x = 5/4 ± √3 = (5±4√3)/4
Since triangle ACD has vertices A,C,D, in the anticlockwise direction, then
D = ((5+4√3)/4, (4+√3)/4)
mark me as brainliest
The square doesn't make sense. I'll assume they meant:
BCEF is a square with vertices B,C,E,F in the clockwise sense.
Besides, there's no way that D can be a vertex of the square with side BC and a vertex of the equilateral triangle with side AC.
A = (1,2)
B = (−3,1)
C = (x,0)
Slope AB = (2−1)/(1+3) = 1/4
Since ABC is right angled at A, then line AB is perpendicular to line AC
Slope AC = −1/(1/4)
−2/(x−1) = −4
1/(x−1) = 2
x − 1 = 1/2
x = 3/2
C = (3/2, 0)
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Since BCEF is a square with vertices B,C,E,F in the clockwise order:
E is point B rotated 270° clockwise (90° counter clockwise) about C
F is point C rotated 270° clockwise (90° counter clockwise) about E
(x,y) rotated 90° counter clockwise about point (a,b)
(x,y) ---> (a+b−y, b−a+x)
E = B(−3,1) rotated 90° ccw about C(3/2, 0)
E = (3/2+0−1, 0−3/2−3) = (1/2, −9/2)
F = C(3/2, 0) rotated 90° ccw about E(1/2, −9/2)
F = (1/2−9/2−0, −9/2−1/2+3/2) = (−4, −7/2)
E = (1/2, −9/2)
F = (−4, −7/2)
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A = (1, 2)
C = (3/2, 0)
D = (x, y)
Since ACD is an equilateral triangle then
AD = CD = AC
AD² = CD² = AC²
(x−1)² + (y−2)² = (x−3/2)² + (y−0)² = (3/2−1)² + (0−2)²
x² − 2x + 1 + y² − 4y + 4 = x² − 3x + 9/4 + y² = 1/4 + 4
x² − 2x + y² − 4y + 5 = x² − 3x + 9/4 + y² = 17/4
x² − 2x + y² − 4y + 5 = x² − 3x + 9/4 + y²
−2x − 4y + 5 = −3x + 9/4
x = 4y − 11/4
x² − 3x + 9/4 + y² = 17/4
(4y − 11/4)² − 3(4y − 11/4) + 9/4 + y² = 17/4
16y² − 22y + 121/16 − 12y + 33/4 + y² − 2 = 0
17y² − 34y + 221/16 = 0
1/16 (16y² − 32y + 13) = 0
y = 1 ± √3/4 = (4±√3)/4
x = 5/4 ± √3 = (5±4√3)/4
Since triangle ACD has vertices A,C,D, in the anticlockwise direction, then
D = ((5+4√3)/4, (4+√3)/4)
mark me as brainliest