50 AMC LECTURES Lecture 2 Analytic Geometry Distance and Lines. can be calculated by the following formula:

Transcription

1 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines BASIC KNOWLEDGE. Distnce formul The distnce (d) between two points P ( x, y) nd P ( x, y) cn be clculted by the following formul: d ( x y () x ) ( y ) Proof: Given right tringle ABC, from the Pythgoren Theorem, we hve AB AC Let AB = d, BC d ( x x ) ( y y Tking the squre root on both sides: d ( x y x ) ( y ) Another useful form of the distnce formul: ) d ( x y = x ) ( y ) x y y = k x x () x ( ) x x Where k is the slope of the line contining the two points nd k y x. y x. Finding the coordintes of point P on AB between points A nd B. AP P is point on AB between points A nd B. If, then we hve the following PB formuls to clculte the coordintes of P: x x y x, nd y y ()

2 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines When =, we get the midpoint formul: The coordintes x, y ) of the midpoint of the line segment with endpoints x, ) nd ( ( m m x, y ) re x x x m, nd y m. Point to line distnce formul ( y y y. (4) (). The distnce from point (x, y ) to line x + by +c = is one of those problems tht seems esy, but is very time consuming, unless you know the following formul: Proof: d Method : x by c (5) b Drop perpendiculrs to the line l from the given point P x, ) to ( y x c x c meet l t N( x, ), to the x-xes to meet l t Q( x, ), b b x c nd to P Q from N to meet P Q t M ( x, ). b The tringle P QN in the picture is right tringle, so the re cn be found in two wys: P Q NM d NQ, or x c y ( ) x b Solving for d: x d x c x c b b x x.

3 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines d y x c ) x b x c x c b b x x c ) x b x c ) x b x x x x x x x x y b x y b x b y ) x c b b x by b c Method : Find the distnce from the point P x, ) to the line l : x by c. ( y l : x by c meets x-xes nd y-xes t c c E(, ), F(, ), respectively. b Drop perpendiculrs to the line l from the given point x c P ( x, y) to meeting l t N( x, ), to the x-xes to meet b x c l t Q( x, ), b So Rt P QN ~ Rt EFO. x c y P N P Q d b x by c d OE EF c c b b b (). Other forms of the formul of point to distnce (i). x by b c x y (6) It tells us tht the distnce from point (x, y ) to the line l: x + by = is less thn or equl to the distnce from the point to origin if the line l psses through the origin nd the point is not on the line.

4 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines Note tht if we squre both sides of (6), we get fmous inequlity: Cuchy Inequlity: ( x by ) ( b )( x ). y (ii). x by b c ( x x) ( y y ) (7) It tells us tht ll the line segments connecting point tht is not on the line l: x + by + c = to point tht is on the line l, the perpendiculr is the shortest. (). If line l x by c nd line l x by c re prllel, the distnce between them is d nd 4. Angle formed by two lines d c c (8) b The slopes of line l nd line l re k nd k, respectively. The ngle formed by lines l nd l is nd tn θ = When k k, =. k k k k (9) The slopes of line l nd line l re k nd k, respectively. The ngle from lines l to l is nd k k tn () k k When k k, =. 5. Pttern in Reflections (). P x, y ) is point. The imge of P under reflections: ( (). In the xxes x, ) ( y (b). In the yxes x, y ) ( 4

5 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines (c). In the x = x, y ) ( (d). In the y = x, ) ( y (e). In the y = x y, ) ( x (f). In the y = x y, ) ( x (g). In the y = x + m y m, x ) ( m (h). In the y = x + n n y, n ) ( x (i). In the point A(, b) x,b ) (j). In the Ax + By +C = The figures for () to (f) re showing below. ( y x x y y A B C ( x, y) : A( y y) B( x x ) () (b) (c) (d) (e) (f) (). The reflection of the line Ax + By +C = in the point P(, b ): Ax + By (A + bb +C) = 5

6 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines Exmple : Find the imge of P( 5,) under the reflection in line x y =. Solution: Let the imge be Q x, ) nd R be the midpoint of PQ. ( y ( 5) 5 PR ( ) 4 So PQ PR. The slope of the line PQ is k PQ. By the formul (), we hve 4 PQ ( ) x 5 Therefore x 5 6 x. We lso hve: y y k PQ y. x 5 6 So the imge Q is (, ). Exmple : A line pssing through (, ) is cut by two prllel lines 4x + y + = nd 4x + y + 6 =. The length of the segment cut is. Find the eqution of the line. Solution: Let k be the slope of the line. Let the line be y k( x ). The two intersecting points with two prllel lines be P x, ) nd P x, ). ( y ( y y k( x ) 4x y k 7 x. k 4 6

7 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines y k( x ) k x. 4x y 6 k 4 k k 7 Since P P k x x, k. k 4 k 4 Simplifying we get: 7k 48k 7 Solve for k: k, 7 7 k. The line seeking is y ( x ) nd y 7( x ). 7 Or x + 7y 5 = nd 7x y 5 =. Exmple : Show tht x y c nd b c if with y bx. x, y,, b, nd c re rel numbers. Solution: Rewrite the given eqution s y bx c ( x ) ( y b) y bx c ( x ) ( y b) This is the distnce from (, ) to the line l : ( b y) x ( x) y y bx. Points (, b) nd (x, y) re on the line. Therefore b c, x y c or b c, x y c. Exmple 4: Find the smllest vlue of u x y x 4y if x y + =. Solution: From u x y x 4y, we hve u 5 ( x ) ( y ). Exmining the point (, ) nd the line x y + = : ( x ) ( y )

8 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines 49 Squring both sides: ( x ) ( y ) Tht is u 5 u The smllest vlue of is chieved when y = nd x =. 5 5 Exmple 5: Line ( m ) x (m ) y m 5lwys contins the point P for ny rel vlue of m. Wht is the coordintes of the point P? Solution: (9, 4). Write the given eqution in the following form: m ( x y ) ( x y 5) For ny rel number of m, the line lwys goes through the point of intersection of lines x y nd x y 5. Solving x nd y: x y x y 5 x = 9 nd y = 4. The nswer is: (9, 4). Exmple 6: If the y-intercept of the line ( m ) x ( m m ) y m is, wht is the vlue of the rel number m? Solution:. Since the y-intercept is, (, ) is on the line. Substitute (, ) into the eqution, we hve: m m m m m Solving for m: m = or m =. If m =, we hve m + = nd ( m ) x ( m m ) y m is not line. So m = is the nswer. Exmple 7: Find the distnce between the point (, ) nd the line x 4y =. Solution: 6/5. 8

9 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines () 4( ) 6 6 By the point to line distnce formul: d Exmple 8: Wht is the closest tht the line 4x + 5y = comes to the origin? Solution: 4() 5() 4 Since 4x + 5y =, we hve d Exmple 9: Find the distnce between the prllel lines x y = nd x y= 6. Solution: Pick ny point on one line nd plug it into the distnce formul for point nd line. So let s pick (6, ) from the first line. Now d (6) () Exmple : Wht is the closest tht the line Duke Mth Meet). 4 y x comes to lttice point? ( 7 5 Solution: The line, in stndrd form, is x 5y + 7 =, so the distnce to ny lttice point (x, y) x 5y 7 is. 5 So the closest point will occur when the numertor is minimized. Since it is n bsolute vlue, nd must be n integer, we need to see if this quntity cn be zero, or if not, wht is the smllest possible vlue. Clerly x 5y will lwys be multiple of 5, so we could mke it 5, mking the entire numertor. This is the smllest possible vlue for 65 the numertor, so the shortest distnce is

10 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines Exmple : P(x, y) is point on circle of rdius of. If the center of the circle is (, y ), wht is the smllest vlue of? x Solution:. 4 y Let k. It is cler tht k is the slope of the line connecting points P(x, y) nd A(, x y ). The smllest vlue of is the smller one of the two slopes of the tngent lines to x the circle. The eqution of the line AP is kx y ( k) By the point to line distnce formul, we hve: k k k k 4 Exmple : Wht is the shortest distnce between the circle x + y = 5 nd the line x + 4y = 48? Solution: 5. () 4() 4 48 First notice tht the closest the line gets to the origin is, so it is 4 5 more thn 5 units from the origin. Now subtrct the rdius of the circle, yielding the 48 distnce Exmple : Find tngent of the ngle formed by two lines both pssing through the origin. It is known tht these two lines trisect the segment in the first qudrnt of the line x + y =. Solution: Let A(6, ) nd B(, 4).

11 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines AC x Since, CB y BD x Since,. DA y C C D D C (, ), D(4, ). k OC, 4 k OD 4 9 tn. 4 Exmple 4: Find the smllest possible vlue of f ( x) x x x 4x 8. Solution: f ( x) ( x ) ( ) ( x ) ( Tht is, to find the smllest possible sum of the distnces from (x, ) to points (, ), (, ). Tht is, the distnce between (, ) nd (, ), which is. )

12 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines PROBLEMS Problem : Wht is the closest tht the line y x 5 7 comes to lttice point? Problem : (966 AMC) Let m be positive integer nd let the lines x + y = 7 nd y = mx intersect in point whose coordintes re integers. Then m cn be: (A) 4 only (B) 5 only (C)6 only (D) 7 only (E) one of the integers 4, 5, 6, 7 nd one other positive integer Problem : (98 AMC) A verticl line divides the tringle with vertices (, ), (,) nd (9,) in the xy-plne into two regions of equl re. The eqution of the line is x = (A).5 (B). (C).5 (D) 4. (E) 4.5 Problem 4: (NC Mth Contest 997) Let P be the point (,). Let Q be the reflection of P bout the x-xis, let R be the reflcetion of Q bout the line y = x nd let S be the reflection of R through the origin. Then PQRS is convex qudrilterl. Wht is the re of PQRS? () 4 (b) 5 (c) 6 (d) 7 (e) 8 Problem 5: (NC Mth Contest ) Find the slope of the line with positive rtionl slope, which psses throug the point(6, ) nd t distnce of 5 from (, ). Write the slope in the form b, where nd b re reltively prime. Wht is the sum of nd b?. 4 b. c. d. e. none of these. Problem 6: Line (l ) x y + = intersects x-xes t A. Line (l ) is obtined by rotting l 45 bout A. If the rottion is counterclockwise, find the eqution for l. Problem 7: Find + b if the line x + y + = is prllel to the line ( + )x + by + = nd two lines hve the sme distnce to the origin (A). (B). (C) or. (D) ny vlue except,. Problem 8: The vertices re A(4, ), B(7, 5), nd C(4, 7) of tringle ABC. Find the eqution of the ngle bisector of A.

13 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines Problem 9: Find the eqution of the line pssing through p(, ) nd the length of its segment cut between two lines 4x y nd 4x y 6 is. Problem : Find the imge of P(, ) under the reflection in line x y. Problem : Find the imge of line x y under the reflection in line x y. Problem : Find the imge of the line x y 4 under the reflection in (, ). Problem : Find the gretest possible vlue of g ( x) x x 5 x 4x. Problem 4: Find the eqution of the line pssing through point P(5, 4). It is known tht P divides the segment A(x, ) nd B(, y) of the line between the x nd y xis into the rtio of :. Problem 5: A line segment is between two lines: l : x y nd l : x y 8. The midpoint of the segment is P(, ). Find the eqution of the line contining the segment. Problem 6: Find the eqution of the ngle bisector of the ngle formed by lines x y nd 7x y 4. Problem 7: Find if the distnce between line l : x + y 6 = nd line l : 4x + 6y + = is 5. 6 Problem 8: Find the eqution of the line pssing through point P(, ) nd hving distnce from the origin. Problem 9: ( AMC ) Points A = (, 9), B = (, ), C = (5, ), nd D = (, b) lie in the first qudrnt nd re the vertices of qudrilterl ABCD. The qudrilterl formed by joining the midpoints of AB, BC, CD, nd DA is squre. Wht is the sum of the coordintes of point D?

14 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines (A) 7 (B) 9 (C) (D) (E) 6 Problem : ( AMC A) A set S of points in the xy-plne is symmetric bout the origin, both coordinte xes, nd the line y = x. If (, ) is in S, wht is the smllest possible number of points in S? (A) (B) (C) 4 (D) 8 (E) 6 Problem : (4 AMC A) Let A = (, 9) nd B = (, ). Points A' nd B ' re on the line y = x, nd AA' nd BB ' intersect t C = (, 8). Wht is the length of AA '? (A) (B) (C) (D) (E) Problem : (4 AMC B) The point (, ) is rotted 9 o clockwise round the origin to point B. Point B is then reflected in the line y = x to point C. Wht re the coordintes of C? (A) (, ) (B) (, ) (C) (, ) (D) (, ) (E) (, ) Problem : ( AMC) The point P = (,, ) is reflected in the xy-plne, then its imge Q is rotted by 8 bout the x-xis to produce R, nd finlly, R is trnslted by 5 units in the positive-y direction to produce S. Wht re the coordintes of S? (A) (, 7, ) (B) (, 7, ) (C) (,, 8) (D) (,, ) (E) (,, ) Problem 4: (98 AMC ) The lines L nd K re symmetric to ech other with respect to the line y = x. If the eqution of line L is y = x + b with nd b, then the eqution of K is y = (A) x b (B) x b (C) x b (D) x b (E) x Problem 5: ( NC Mth Contest Algebr II) Given A(,), B(5, ), nd C(, ), find the eqution of the ngle bisector t A. b 4

15 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines. 5x y 7 b. 7x y c. 7y x d. y x e. none of the bove Problem 6: (98 AMC ) The equtions of L nd L re y = mx nd y = nx, respectively. Suppose L mkes twice s lrge n ngle with the horizontl (mesured counterclockwise from the positive x- xis) s does L, nd tht L hs 4 times the slope of L. If L is not horizontl, then mn is (A) (B) (C) (D) (E) not uniquely determined by the given informtion Problem 7: (99 AIME) A tringle hs vertices P = ( 8, 5), Q = ( 5, 9) nd R = (, 7). The eqution of the bisector of P cn be written in the form x + y + c =. Find + c. 5

16 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines SOLUTIONS TO PROBLEMS Problem : Solution: Let m nd n re the coordintes of lttice point. The distnce from this lttice point (m, n) to the given line is d m n m 5n 7 4 7(m 5n) 7 4 We know tht 7(m 5n) is multiple of 7. We lso know tht nd 5 re reltively prime. Since we wnt to get the smllest vlue for d, or the smllest vlue for the numertor, so 7(m 5n) = 7 is the best vlue we could get (m = nd n = ). d min Problem : Solution: (C). Substituting y from the second eqution into the first gives x + (mx ) = 7, so tht 7 79 x. m m Since x is to be n integer, the denomintor + m must be divisor of the numertor, nd its only divisors re,,, 79, 79, 79. Our tsk now is to find positive integer m such tht + m = d, or m d, Where d is one of these divisors. Since m >, we see tht d >, so the only divisore we need to test re the lst three: 66 (i) if d = 79, d = 66, nd m 6 (ii) if d = 79 = 7, d = 4 is not divisible by (iii) if d = 79 = 7, d = 698 is not divisible by. We conclude tht m = 6 is the only positive integer yielding lttice point for the intersection of the given lines. 6

17 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines Problem : Solution:(B). Method (officil solution): In the djoining figure, ABC is the given tringle nd x = is the dividing line. Since re ABC ()(8) 4, the two regions must ech hve re. Since the portion of ABC to the left of the verticl line through vertex A hs re less thn re ABF, x the line x = is indeed right of A s shown. Since the eqution of line BC is y, the 9 verticl line x = intersects BC t point E: (, ). Thus 9 re DEC (9 ), or (9 ) = 6. 9 Then 9 = ± 6, nd = 5 or. Since the line x = must intersect ABC, x =. Method (our solution): Extend CA to meet y-xes t D. We see tht CDA CGE. Therefore, we get: DB CD 9 CG GE GE CG GE CG 9 We lso see tht the re of CDA = the re of CAB + the re of t DAB = the re of CGE + the re of t DAB. CD DB CG GE AD DB 9 CG GE CG GE 4 CG 4 CG 9 CG 6 CG 6. DG = DC CG = 9 6 =. Therefore x =. Problem 4: Solution: The points P, Q, R, nd S re respectively (, ), (, ), (, ) nd (, ). The 7

18 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines rectngle with vertices(, ), (, ), (, ) nd (, ) hs re. When the three corners, with, 5 nd re subtrcted, the remining qudrilterl hs re5. Problem 5: Solution: b. 8 DE 5, BE 4, OC. 5 DBE z DBO x 5 tn z tn( x y). 8 5 tn z tn y 8 tn y. tn x. 6 5 tn z tn y 5 Answer: =. Problem 6: Solution: Let the slope of l be k nd l be. tn We hve tn nd k tn(45 ). tn Since the intersecting point is A(, ), the eqution of line is then y = (x + ) or x y + 9 =. Problem 7: Solution: (A). Since two lines re prllel, b = + () Since they hve the sme distnce to the origin, () ( ) b Solving the system of equtions () nd () gives: or b b. When =, b =, these two lines re overlpped. 8

19 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines 7 Therefore, b = b. Problem 8: Solution: Let p(x, y) be point on the ngle bisector AD. Method : The distnce from p to AC is the sme s the distnce from p to AB. 4 The eqution for AB: y ( x 4) or 4x y. 4 The eqution for AC: y ( x 4) or x 4y 6. 4x y 4 ( ) x 4y 6 4 4x y ( x 4y 6) x 7y or 7x y 9. It is esy to know tht x 7y is the eqution of the exterior ngle of A. The eqution is then 7x y 9. Method : By the ngle bisector theorem, So BD DC D x, then: x x The eqution is: BD DC 7 ( 4) AB AC, nd (4 7) (4 4) D y ( 5) ( 7) y y

20 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines 7 y y y y y 7 x x x x x 4 4 7x y 9 y 7x 4 Problem 9: Solution: Method : Let the eqution be y k( x ). The points of intersection with two lines re A nd B, respectively. Solving the system equtions of 4x y y k( x ) nd 4x y 6 4x y The coordintes of A nd B re: k 7 5k 8 k 8 k (, ) nd (, ). k 4 k 4 k 4 k 4 Since AB 5 5k ( ) ( ) k 4 k 4, 5 k k 4. The bove eqution hs the form of: (7k + )(k 7) = So the eqution is x y 5 or 7x y 5. k or 7. 7 Method : The distnce between two prllel lines is d 6 4. Since the length of the segment cut by them is, the ngle formed the line in question nd two prllel lines is 45.

21 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines Let the eqution be y k( x ). tn k k k k 4 k 4 k Solving for k: k, 7 7 k. So the eqution is x y 5 or 7x y 5. Problem : Solution: Let the imge be P (x, y). x y x y 8 y ( ) x y x x 5 ' 9 Solve for x nd y: P (, ) y. 5 Problem : Solution: Method : x y Solving x y 5 9 gives the point of intersection: (, ) Let the slope of the line in question be k: k k Solve for k: k = 7 or k = (extrneous). 9 5 So the eqution of the line is y 7 ( x ) or 7x y. Method : Since P (, ) is point on x y, its imge is P ( x, y).

22 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines y x x y x x Solve for x: or (extrneous). y y 9 So k The eqution is y 7 ( x ). Problem : Solution: Let the eqution we wnt to find be y x b We know tht d, the distnces from (, ) to these two prllel lines re the sme. 4 d. b Therefore Solving for b: b = or b = 4 (extrneous). So the eqution is y = x. Problem : Solution: g ( x) ( x ) ( x ) To find the gretest difference PA PB of the distnces from P to A (, ) nd P to B(, ). The desired point P must be the point of intersection of the line segment connecting A nd B nd x-xes. AB ( ) ( )

23 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines Problem 4: Solution: x y Let the eqution of the line be:. The coordintes of A nd B re A(, ), B(, b b.). AP 5 Since, considering P(5, 4), by formul (), we obtin:, b =. PB The eqution is 8x 5y 6. Problem 5: Solution: Let the eqution be y = kx +. The points of intersection of line y = kx + with 7 4k 7 8k x y nd x y 8 re A(, ) nd B (, ), k k k k respectively. By the midpoint formul (4), we get k. 4 The eqution is y x or x 4y 4. 4 Problem 6: Solution: Let P(x, y) be point on the ngle bisector. The distnces from point P to two given lines re the sme. So x y 7x y 4. 5 The equtions re: 6x y nd x y 7. Problem 7: Solution: 5 We know tht the distnce between l : x + y 6 = nd l : l : x + y + = is. 6 By the formul (8) d 5 Therefore 6 C C A B, we hve: Solving for gives = 7 or 7..

24 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines Problem 8: Solution: If the slope of the line does not exist, the eqution of the line is x =. If the slope of the eqution exists, let it be k. The eqution is then written s y + = k(x ), or kx y k =. k By the point to line distnce formul, we hve =. k Solve for k: k. 4 The eqution of the line is: y ( x ) or x 4y =. 4 The equtions re x = or x 4y =. Problem 9: Solution: (C). Let the midpoints of sides AB, BC, CD, nd DAbe M, N, P, nd Q, respectively. Then M = (, 5) nd N = (, ). Since MN hs slope, the slope of MQ must be, nd MQ ( x ) = MN =. An eqution for the line contining MQ is thus y 5, or ( ) x ( ) y. Therefore Q hs coordintes of the form (, ). Since MQ, we ( ) hve ( ) ( 5) ( ) ( ) ( ) 9 ( ) 9. We know tht Q is in the first qudrnt, so = 5 nd Q = (5, 6). Since Q is the midpoint of AD nd A = (, 9), we hve D = (7, ), nd the sum of the coordintes of D is. Problem : Solution: (D). The set S is symmetric bout the line y = x nd contins (, ), so it must lso contin (, ). Also S is symmetric bout the x-xis, so it must contin (, ) nd (, ). Finlly, since S is symmetric bout the y-xis, it must contin (, ), (, ), (, ), nd (, ). Since the resulting set of 8 points is symmetric bout both coordinte xes, it is lso symmetric bout the origin. 4

25 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines Problem : Solution: (B). Line AC hs slope nd y-intercept (, 9), so its eqution is y x 9. Since the coordintes of A stisfy both this eqution nd y = x, it follows tht A = (6, 6). Similrly, line BC hs eqution y = x +, nd B = (4, 4). Thus A B (6 4) (6 4). Problem : Solution: (E). The rottion tkes (, ) into B = (, ), nd the reflection tkes B into C = (, ). Problem : Solution: (E). Reflecting the point (,, ) in the xy-plne produces (,, ). A hlf-turn bout the x- xis yields (,, ). Finlly, the trnsltion gives (,, ). Problem 4: Solution: (E). If (p, q) is point on line L, then by symmetry (q, p) must be point on K. Therefore, the points on K stisfy x = y + b. x b Solving for y yields y. Problem 5: Solution: (e). We find tht side AC nd AB both hve length 5, so the ngle bisector is lso the medin, pssing through A(,) nd the midpoint (/,/), of side BC. This mkes the slope 7 nd the eqution y 7(x ) 7x

26 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines Problem 6: Solution: (C). In the djoining figure, L nd L intersect the line x = t B nd A, respectively; C is the intersection of the line x = with the x-xis. Since OC =, AC is the slope of L nd BC is the slop of L. Therefore AC = n. OC AC Since OA is n ngle bisector,. OB AB n This yields nd OB =. OB n By the Pythgoren theorem (4n) 9, so n. Problem 7: Solution: Method : Let the bisector be PT nd the coordintes of T be (x, y). PQ 7 4 5, PR 9 5. QT TR PQ PR By formul (), we hve x = 5, y. The eqution for PT: y 5 ( x 8) or x + y + 78 =. Therefore + c = + 78 = 89. Method : Let the slope of the ngle bisector be k. The slopes of PR nd PQ re k PR nd k PQ, respectively k PR k PQ By the formul (), we hve: 4 4 k k 7 k + 7k = k, k. 4 4 k k 7 6

27 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines Since the ngle bisector is the interior ngle bisector, y 5 ( x 8) or x + y + 78 = Therefore + c = + 78 = 89. k. 7

28 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines EXTRA EXERCISES Exercise : In the xy- plne, how mny lines whose x-intercept is positive prime number nd whose y-intercept is positive integer pss through the point (4, )? (994 AMC ). Answer:. Exercise : If the line whose x-intercept nd y-intercept hve the smllest sum psses through the point P(, 4), find the slope of the line. It is known tht both the x-intercept nd y-intercept re positive. Answer:. Exercise : Find the distnce between the point (, ) nd the line x y + =. Answer:. Exercise 4: Wht is the positive vlue of m if line with the slope of pssing through the point (, m) nd is tngent to the circle x + y =? Answer:. Exercise 5: Wht is the positive slope of the line tht goes through the point (, ) nd is tngent to the circle x + y =? Express your nswer in the simplest rdicl form. Answer:. Exercise 6: P(x, y) is point on circle of rdius of. If the center of the circle is (, ), wht is the sum of the smllest nd the gretest vlues of x y? Answer: 4. Exercise 7: P(x, y) is point on circle of rdius of. If the center of the circle is (, ), wht is the lrgest vlue of x y? 8

29 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines Answer:. Exercise 8: Among ll lines pssing through the point P(, 5), line l hs the gretest distnce from the origin. Find the eqution of the line l. Answer: x + 5y 4 = Exercise 9: The re of the tringle bounded by the x-xis, the y-xis nd the line x y + k = is. Wht is the gretest vlue of k? Answer: k =. Problem : The re of the tringle bounded by the x-xis, the y-xis nd the line l hs the smllest possible vlue. Wht is the slope of the line l? Answer: /. Exercise : Find the eqution of the imge of the line x y rotting counterclockwise with respect to point P(, ). Answer: x y. Exercise : Find the eqution of the line pssing through point P(, ). It is known tht the ngle formed by this line nd line x y = is 45. (A) y x + 5 = or y + x = ; (B) y x + 5 = or y x + 4 = ; (C) y + x + = or y + x = ; (D) y + x + = or y x + 4 =. Answer: (A) y x + 5 = or y + x =. Exercise : Find the rnge for if the distnce from point (4, ) to line 4x y = is less thn or equl to. (A) (, ) (B) (, ); (C) (, ); (D) (, ) (, +) Answer: (C). Exercise : Find if the distnce from point P(, ) to line x + ( )y+ = is. 9

30 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines Answer: 7 or. Exercise 4: Find if the distnce from point (, ) to line x 4y 4 is. Answer: 7 or 7 Exercise 5. Find the eqution of the line pssing through point P(, ) nd hving n ngle of 45 to line x 5y. Answer: x 7y 7 Exercise 6: Find the eqution of the line pssing through the intersecting point of lines x y 5 nd x y 5. The line nd line x y hve n ngle of. 5 Answer: y or y x. Exercise 7: Find the coordintes of the imge of P(, b) under the reflection in line x y + =. (A) (b, + ); (B) (b +, ); (C) (, b + ); (D) ( +, b ). Answer: (A). Exercise 8: Find the eqution of the imge of lx + y 4 = under the reflection in line x =. (A) x y + = ; (B) x + y 6 = ; (C) x + y 8 = ; (D) x y =. Answer: D. Exercise 9: Find nd b if point A( +, b + ) is the imge of the point B( +, b + ) under the reflection in line 4x + y =. Answer: = 4 nd b =.

31 5 AMC LECTURES Lecture Anlytic Geometry Distnce nd Lines Exercise : Find the eqution of the imge of x y + = under the reflection in line x + y =. Answer: x y + 4 =. Exercise : Find the eqution of the imge of x y + = under the reflection in point A(,). Answer: x y 5 =.