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

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.

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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