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1. Find the distance from the point to the given line by using the distance theorem http://mypages.iit.edu/~maslanka/PointtoLine.pdf ( a ) 2 x– 4 y + 2 = 0 ; (1,3) (b) 4 x + 5 y – 3= 0 ; (-2,4)

1. Find the distance from the point to the given line by using the distance theorem http://mypages.iit.edu/~maslanka/PointtoLine.pdf ( a ) 2 x– 4 y + 2 = 0 ; (1,3) (b) 4 x + 5 y – 3= 0 ; (-2,4)

Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
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See question #1 in handout along with formula. Find the distance to the given line using distance theorem. 

signment 11 Due date : Friday, December 11 Section 22.4, Page 707 #1, 2, 6*, 7*, 14, 15, 17. *Note: The vertex is at the origin in these problems. Required Additional Exercises Exercises #1 – 5 (below). 1. Find the distance from the point to the given line by using the distance theorem http://mypages.iit.edu/~maslanka/PointtoLine.pdf (a) 2 x – 4 y + 2 = 0 ; (1,3) (b) 4х+5у-3-0; (-2,4) 2. Find the distance between the parallel lines: L: 2 x- 5 y +3 = 0; L2: 2 x – 5 y + 7 = 0. (Hint: It suffices to compute the distance from any particular point P, on L to the line La.) 3. Find the equation of the line L bisecting the angle from Li to La given Li: 3 x – 4 y – 2 = 0 ; Hint: Note that if P = ( x , y) is any point on the bisector L then its coordinates satisfy L2: 4 x – 3 y + 4 = 0. the condition: |A,x + B1y+ C¡| _ |A,x + B,y+ C¿| VA,? + B,2 VA,² + B,² |A,² + B,² · ( Aµx + B¡y+ C;) = ± VA,² + B,² · (A,x + B2y+ C,) (*) where L: A,x+B,y +C, =o and L2: A,x + B,y + C, = 0 . The solutions to (*) yield the equations of both lines bisecting the angles between L, and L2. 4. Find the two points of intersection of the circles: x² + y 2 + 5 x + y – 26 = 0 ; x² + y² + 2 x – y – 15 = 0. 5. Find the equation of the parabola P with focus F = (1,1) and the directrix D: x + y = 0 in two ways: (a) By using the fact that : Q = ( x , y ) on P → d(Q,F ) = d ( Q , D ). Hint: Square both sides of your equation for the parabola in order to eliminate the square roots and absolute values and then simplify it. (b) By rotating the parabola y² =2 z x by 45° about its axis and translating 1 its graph + units horizontally and +; units vertically. Hint: Transform the equation of the parabola to polar coordinates before rotating this curve. Then transfer back to xy-coordinates in order to translate the equation of the rotated parabola. Refer to the handout: http://mypages.iit.edu/~maslanka/R&T_Thrms.pdf for details on rotation and translations theorems.
Transcribed Image Text:signment 11 Due date : Friday, December 11 Section 22.4, Page 707 #1, 2, 6*, 7*, 14, 15, 17. *Note: The vertex is at the origin in these problems. Required Additional Exercises Exercises #1 – 5 (below). 1. Find the distance from the point to the given line by using the distance theorem http://mypages.iit.edu/~maslanka/PointtoLine.pdf (a) 2 x – 4 y + 2 = 0 ; (1,3) (b) 4х+5у-3-0; (-2,4) 2. Find the distance between the parallel lines: L: 2 x- 5 y +3 = 0; L2: 2 x – 5 y + 7 = 0. (Hint: It suffices to compute the distance from any particular point P, on L to the line La.) 3. Find the equation of the line L bisecting the angle from Li to La given Li: 3 x – 4 y – 2 = 0 ; Hint: Note that if P = ( x , y) is any point on the bisector L then its coordinates satisfy L2: 4 x – 3 y + 4 = 0. the condition: |A,x + B1y+ C¡| _ |A,x + B,y+ C¿| VA,? + B,2 VA,² + B,² |A,² + B,² · ( Aµx + B¡y+ C;) = ± VA,² + B,² · (A,x + B2y+ C,) (*) where L: A,x+B,y +C, =o and L2: A,x + B,y + C, = 0 . The solutions to (*) yield the equations of both lines bisecting the angles between L, and L2. 4. Find the two points of intersection of the circles: x² + y 2 + 5 x + y – 26 = 0 ; x² + y² + 2 x – y – 15 = 0. 5. Find the equation of the parabola P with focus F = (1,1) and the directrix D: x + y = 0 in two ways: (a) By using the fact that : Q = ( x , y ) on P → d(Q,F ) = d ( Q , D ). Hint: Square both sides of your equation for the parabola in order to eliminate the square roots and absolute values and then simplify it. (b) By rotating the parabola y² =2 z x by 45° about its axis and translating 1 its graph + units horizontally and +; units vertically. Hint: Transform the equation of the parabola to polar coordinates before rotating this curve. Then transfer back to xy-coordinates in order to translate the equation of the rotated parabola. Refer to the handout: http://mypages.iit.edu/~maslanka/R&T_Thrms.pdf for details on rotation and translations theorems.
THEOREM: The distance from the point P, = ( xo , y. ) to the line l: Ax + By + C = 0 is: | Ax, +By, +C| VA2 +B2 d = Proof: We seek the length of the segment |P,P1l where P.P11l. It is simpler to compute |P,P11? = (x, – X1)² + (yo - Y1)² ( * ) where (x1 , y1) are the coordinates of the yet unkr A mį = --= MpP, B В Observe that A Yo -Y1 В so Xo -x1 А B (x, - X1) A Yo - Y1 = (1) and since P, lies on l it satisfies the condition: Ах1 + Вy, + C %3D 0
Transcribed Image Text:THEOREM: The distance from the point P, = ( xo , y. ) to the line l: Ax + By + C = 0 is: | Ax, +By, +C| VA2 +B2 d = Proof: We seek the length of the segment |P,P1l where P.P11l. It is simpler to compute |P,P11? = (x, – X1)² + (yo - Y1)² ( * ) where (x1 , y1) are the coordinates of the yet unkr A mį = --= MpP, B В Observe that A Yo -Y1 В so Xo -x1 А B (x, - X1) A Yo - Y1 = (1) and since P, lies on l it satisfies the condition: Ах1 + Вy, + C %3D 0
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