Find: R [ft] y [ft] C (5,6) B (2,5) 4.6 5.1 5.8 6.4 A (0,0) x [ft] Find the radius of the curve, in feet. [pause] In this problem, points A, B and C ---- A (0,0) x [ft]
Find: R [ft] y [ft] C (5,6) B (2,5) 4.6 5.1 5.8 6.4 A (0,0) x [ft] are plotted on a x y plane, where the coordinates --- A (0,0) x [ft]
Find: R [ft] y [ft] C (5,6) B (2,5) 4.6 5.1 5.8 6.4 A (0,0) x [ft] of all 3 points are provided. [pause] If a circular curve is drawn --- A (0,0) x [ft]
Find: R [ft] y [ft] C (5,6) B (2,5) 4.6 5.1 5.8 6.4 A (0,0) x [ft] connecting these three points, the problem asks, what is the radius --- A (0,0) x [ft]
Find: R [ft] C (5,6) y [ft] B (2,5) O (Ox,Oy) R A (0,0) x [ft] of a such a curve. We’ll equate the radius to the distance between points ---- A (0,0) x [ft]
Find: R [ft] C (5,6) y [ft] B (2,5) R= (Ox-Ax)2+(Oy-Ay)2 O (Ox,Oy) R A and O. Since point A is at the origin, the equation for the radius ---- A (0,0) x [ft]
Find: R [ft] C (5,6) y [ft] B (2,5) R= (Ox-Ax)2+(Oy-Ay)2 R= Ox2 + Oy2 O (Ox,Oy) R simplifies to the root sum squared of the x and y coordinates of point O, --- A (0,0) x [ft]
Find: R [ft] C (5,6) y [ft] B (2,5) R= (Ox-Ax)2+(Oy-Ay)2 R= Ox2 + Oy2 O (Ox,Oy) R which is a point defined at the center of the curve. We’ll find the coordinates of Point O --- A (0,0) x [ft]
Find: R [ft] R= Ox2 + Oy2 C (5,6) y [ft] B (2,5) O (Ox,Oy) R A (0,0) by finding the intersection of lines, D O, and --- A (0,0) x [ft]
Find: R [ft] R= Ox2 + Oy2 C (5,6) y [ft] B (2,5) E D O (Ox,Oy) R and E O. We can define Line D O by ---- A (0,0) x
Find: R [ft] R= Ox2 + Oy2 C (5,6) y [ft] B (2,5) E D O (Ox,Oy) R drawing Chord A B, and then drawing a line normal to Chord A B which ---- A (0,0) x
Find: R [ft] R= Ox2 + Oy2 C (5,6) y [ft] B (2,5) E D O (Ox,Oy) R passes through the midpoint of Chord A B, which occurs at Point D. Line E O is generated in a similar fashion, ---- A (0,0) x
Find: R [ft] R= Ox2 + Oy2 C (5,6) y [ft] B (2,5) E D O (Ox,Oy) R Chord B C is drawn out, and line E O --- A (0,0) x
Find: R [ft] R= Ox2 + Oy2 C (5,6) y [ft] B (2,5) E D O (Ox,Oy) R bisects and is perpendicular to Chord B C. Using geometry, --- A (0,0) x
Find: R [ft] R= Ox2 + Oy2 C (5,6) y [ft] B (2,5) E D O (Ox,Oy) R we can show how the center of the curve, Point O, must coincide at --- A (0,0) x
Find: R [ft] R= Ox2 + Oy2 C (5,6) y [ft] B (2,5) E D O (Ox,Oy) R the intersection of lines D O and E O. Both Triangle A B O and ---- A (0,0) x
Find: R [ft] R= Ox2 + Oy2 C (5,6) y [ft] B (2,5) E D O (Ox,Oy) A (0,0) Triangle B C O are isosceles, since line segments A O, B O and --- A (0,0) x
Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) E R R D x R O (Ox,Oy) C O all equal to the radius of the curve. Then, from the law of lines, --- R O (Ox,Oy) A (0,0)
Find: R [ft] = = R= Ox2 + Oy2 C (5,6) y B (2,5) sin(ABAO) sin(AABO) E LBO LAO R R D sin(ACBO) sin(ABCO) = LCO LBO x if we substitute in the radius, for the lengths A O, B O and C O, --- R O (Ox,Oy) A (0,0)
Find: R [ft] = = R= Ox2 + Oy2 C (5,6) y B (2,5) sin(ABAO) sin(AABO) E LBO LAO R R D sin(ACBO) sin(ABCO) = LCO LBO x the denominators and sine terms drop out, such that, Angle B A O equals --- R O (Ox,Oy) A (0,0)
Find: R [ft] = = R= Ox2 + Oy2 C (5,6) y B (2,5) sin(ABAO) sin(AABO) E LBO LAO R R ABAO=AABO D sin(ACBO) sin(ABCO) = LCO LBO x Angle A B O, and Angle C B O equals Angle B C O. For these 2 pairs of angles to be equal, --- ACBO=ABCO R O (Ox,Oy) A (0,0)
Find: R [ft] = = R= Ox2 + Oy2 C (5,6) y B (2,5) sin(ABAO) sin(AABO) E LBO LAO ABAO=AABO D sin(ACBO) sin(ABCO) = LCO LBO x Point O, must lie on the lines which bisect, and are perpendicular to, Chords A B and B C. If Point O lies not on either Line D O or Line E O, --- O ACBO=ABCO A (0,0)
Find: R [ft] = = R= Ox2 + Oy2 C (5,6) y B (2,5) sin(ABAO) sin(AABO) E LBO LAO ABAO=AABO D sin(ACBO) sin(ABCO) = LCO LBO x this would violate the angle equalities, which we know --- ACBO=ABCO A (0,0) O
Find: R [ft] = = R= Ox2 + Oy2 C (5,6) y B (2,5) sin(ABAO) sin(AABO) E LBO LAO ABAO=AABO D sin(ACBO) sin(ABCO) = LCO LBO x to be true based on geometry. [pause] Therefore, all we have to do is --- ACBO=ABCO A (0,0) O
Find: R [ft] = = R= Ox2 + Oy2 C (5,6) y B (2,5) sin(ABAO) sin(AABO) E LBO LAO ABAO=AABO D sin(ACBO) sin(ABCO) = LCO LBO x solve for the point where lines D O and E O intersect. The general equation for a line in 2 dimensions is --- O ACBO=ABCO A (0,0) line DO line EO
Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) y-yo = m * (x-xo) E D x O y minus y not equals m times the quantity x minus x not. In this equation, x not and y not --- O A (0,0) line DO line EO
Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) y-yo = m * (x-xo) E known coordinates D of a point on the line x are known coordinates of a point on the line. So for line D O, we’ll determine ---- O line DO A (0,0) line EO y-yo = m * (x-xo) y-yo = m * (x-xo)
Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) y-yo = m * (x-xo) E known coordinates D of a point on the line x the x and y coordinates of Point D, and then substitute those values in for ---- O line DO A (0,0) line EO y-yo = m * (x-xo) y-yo = m * (x-xo)
Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) y-yo = m * (x-xo) E known coordinates D of a point on the line x x not and y not. For line E O, we’ll use the x and y coordinates for point E --- O A (0,0) line DO line EO y-Dy = m * (x-Dx) y-yo = m * (x-xo)
Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) y-yo = m * (x-xo) E known coordinates D of a point on the line x [pause] The variable m represents the slope of the line, --- O A (0,0) line DO line EO y-Dy = m * (x-Dx) y-Ey = m * (x-Ex)
Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) y-yo = m * (x-xo) E slope known coordinates D of a point on the line x in y over x. To the equations for line D O and line E O, we’ll add a subscript --- O A (0,0) line DO line EO y-Dy = m * (x-Dx) y-Ey = m * (x-Ex)
Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) y-yo = m * (x-xo) E slope known coordinates D of a point on the line x D O and E O, to their slope terms. [pause] Next we’ll find the coordinates of Points D and E, --- O A (0,0) line DO line EO y-Dy = mDO * (x-Dx) y-Ey = mEO * (x-Ex)
Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) Ax+Bx Ay+By Dx= Dy= E 2 2 Bx+Cx By+Cy D Ex= Ey= 2 2 by averaging the given coordinates of Points A and B, and Points B and C, respectively. O A (0,0) line DO line EO y-Dy = mDO * (x-Dx) y-Ey = mEO * (x-Ex)
Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) Ax+Bx Ay+By Dx= Dy= E 2 2 Bx+Cx By+Cy D Ex= Ey= 2 2 After plugging in the coordinates, we know point D is located at --- O A (0,0) line DO line EO y-Dy = mDO * (x-Dx) y-Ey = mEO * (x-Ex)
Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) 0+2 0+5 Dx= Dy= E 2 2 Dx=1 2+5 5+6 D Ex= Ey= 2 2 x equals 1, y equals 2.5 and Point E is located at x equals 3.5, y equals 5.5. After plugging in these coordinates, --- Ex=3.5 Ey=5.5 O A (0,0) line DO line EO y-Dy = mDO * (x-Dx) y-Ey = mEO * (x-Ex)
Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) 0+2 0+5 Dx= Dy= E 2 2 Dx=1 2+5 5+6 D Ex= Ey= 2 2 we still have to solve for the slopes of lines D O and --- Ex=3.5 Ey=5.5 O A (0,0) line DO line EO y-2.5 = mDO * (x-1) y-5.5= mEO * (x-3.5)
? ? Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) 0+2 0+5 Dx= Dy= E 2 2 2+5 5+6 D Ex= Ey= 2 2 E O. First we’ll find the slope of lines A B and ---- Ex=3.5 Ey=5.5 O A (0,0) ? line DO ? line EO y-2.5 = mDO * (x-1) y-5.5= mEO * (x-3.5)
? ? Find: R [ft] R= Ox2 + Oy2 C (5,6) y (By-Ay) B (2,5) mAB = E (Bx-Ax) (Cy-By) mBC = (Cx-Bx) D B C. Then, we’ll determine the slopes of lines D O and E O since the slopes of ---- O A (0,0) ? ? y-2.5 = mDO * (x-1) y-5.5= mEO * (x-3.5)
Find: R [ft] R= Ox2 + Oy2 C (5,6) y (By-Ay) B (2,5) mAB = E (Bx-Ax) mBC = (Cx-Bx) D mDO =-mAB-1 perpendicular lines in 2 dimensions are negative reciprocals of each other. After substituting in --- O mEO =-mBC-1 A (0,0) y-2.5 = mDO * (x-1) y-5.5= mEO * (x-3.5)
Find: R [ft] R= Ox2 + Oy2 C (5,6) y (By-Ay) B (2,5) mAB = E (Bx-Ax) mBC = (Cx-Bx) D mDO =-mAB-1 the coordinates of Points A, B and C, we find the slope of line A B equals 2.5, --- O mEO =-mBC-1 A (0,0) y-2.5 = mDO * (x-1) y-5.5= mEO * (x-3.5)
Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) (5-0) mAB = =2.5 E (2-0) (6-5) mBC = =0.333 (5-2) D mDO =-mAB-1 and the slope of Line B C equals, 0.333. After taking the negative reciprocal of these values --- O mEO =-mBC-1 A (0,0) y-2.5 = mDO * (x-1) y-5.5= mEO * (x-3.5)
Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) (5-0) mAB = =2.5 E (2-0) (6-5) mBC = =0.333 (5-2) D mDO =-mAB-1 we find the slopes of lines D O and E O, which equal, --- O mEO =-mBC-1 A (0,0) y-2.5 = mDO * (x-1) y-5.5= mEO * (x-3.5)
Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) (5-0) mAB = =2.5 E (2-0) (6-5) mBC = =0.333 (5-2) D mDO =-mAB-1 =-0.4 negative 0.4 and negative 3. Now we have the equations for line D O and line E O, which simplify to --- O mEO =-mBC-1 =-3 A (0,0) y-2.5 = mDO * (x-1) y-5.5= mEO * (x-3.5)
Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) 0.4 * x + y = 2.9 E mDO =-2.5-1 =-0.4 0.4 times x plus y equals 2.9, and 3 times x plus y equals 16. Subtracting the first equation from the second, --- O mEO =-0.333-1 =-3 A (0,0) y-2.5 = -0.4 * (x-1) y-5.5= -3 * (x-3.5)
Find: R [ft] +( ) R= Ox2 + Oy2 C (5,6) y B (2,5) -1*( ) -1*( ) 0.4 * x + y = 2.9 E +( ) 3 * x + y = 16.0 2.6 * x = 13.1 D x = 5.04 we find x equals 5.04. Substituting 5.04 in for x, we can solve for y, which equals --- O A (0,0) y-2.5 = -0.4 * (x-1) y-5.5= -3 * (x-3.5)
Find: R [ft] +( ) R= Ox2 + Oy2 C (5,6) y B (2,5) -1*( ) -1*( ) 0.4 * x + y = 2.9 E +( ) 3 * x + y = 16.0 2.6 * x = 13.1 D x = 5.04 0.88. 5.04 and 0.88 correspond to the x and y coordinates of Point O, --- O A (0,0) y=0.88 y-2.5 = -0.4 * (x-1) y-5.5= -3 * (x-3.5)
Find: R [ft] +( ) R= Ox2 + Oy2 C (5,6) y B (2,5) -1*( ) -1*( ) 0.4 * x + y = 2.9 E +( ) 3 * x + y = 16.0 2.6 * x = 13.1 D x = 5.04 the center of the curve. Plugging these values into our equation --- O (5.04,0.88) A (0,0) y=0.88 y-2.5 = -0.4 * (x-1) y-5.5= -3 * (x-3.5)
Find: R [ft] +( ) R= Ox2 + Oy2 C (5,6) y B (2,5) -1*( ) -1*( ) 0.4 * x + y = 2.9 E +( ) 3 * x + y = 16.0 2.6 * x = 13.1 D x = 5.04 for the radius, the radius of the curve equals, --- O (5.04,0.88) A (0,0) y=0.88 y-2.5 = -0.4 * (x-1) y-5.5= -3 * (x-3.5)
Find: R [ft] +( ) R= Ox2 + Oy2 C (5,6) y R= 5.12 [ft] B (2,5) -1*( ) -1*( ) 0.4 * x + y = 2.9 E +( ) 3 * x + y = 16.0 2.6 * x = 13.1 D x = 5.04 5.12 feet. [pause] O (5.04,0.88) A (0,0) y=0.88 y-2.5 = -0.4 * (x-1) y-5.5= -3 * (x-3.5)
Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) R= 5.12 [ft] E 4.6 5.1 5.8 6.4 D When reviewing the possible solutions, --- O (5.04,0.88) A (0,0) y-2.5 = -0.4 * (x-1) y-5.5= -3 * (x-3.5)
Find: R [ft] R= Ox2 + Oy2 C (5,6) y B (2,5) R= 5.12 [ft] E AnswerB 4.6 5.1 5.8 6.4 D the answer is B. O (5.04,0.88) A (0,0) y-2.5 = -0.4 * (x-1) y-5.5= -3 * (x-3.5)
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