# 10.3 Coordinate Proof Using Distance with Segments and Triangles

## Transcription

10.3 Coordinate Proof Using Distance with Segments and Triangles

DO NOT EDIT--Changes must be made through "File info" CorrectionKey=NL-A;CA-A Name Class Date 10.3 Coordinate Proof Using Distance with Segments and Triangles Resource Locker Essential Question: How do you write a coordinate proof? Deriving the Distance Formula and the Midpoint Formula Explore Complete the following steps to derive the Distance Formula and the Midpoint Formula. A To derive the Distance Formula, start with points J and K as shown in the figure. K (x2, y2) y Given: J(x 1, y 1) and K(x 2, y 2) with x 1 ≠ x 2 and y 1 ≠ y 2 Prove: JK = ―――――――― √(x 2 - x 1) 2 + (y 2 - y 1) 2 L x J (x1, y1) _ Locate point L so that JK is the hypotenuse of right triangle JKL. What are the coordinates of L? (x 2, y 1) B Find JL and LK. JL = x 2 - x 1, LK = y 2 - y 1 © Houghton Mifflin Harcourt Publishing Company C By the Pythagorean Theorem, JK 2 = JL 2 + LK 2. Use this to find JK. Explain your steps. JK = (x 2 - x 1) 2 + (y 2 - y 1) 2 by substitution. 2 Taking the square root of both sides shows that JK = D ―――――――― √(x 2 - x 1) 2 + (y 2 - y 1) 2 . To derive the Midpoint Formula, start with points A and B as shown in the figure. Given: A(x 1, y 1) and B(x 2, y 2) _ y1 + y2 x1 + x2 _ , Prove: The midpoint of AB is M _ . 2 2 ( ) y B (x2, y2) y2 M y1 A (x1, y1) x1 x x2 What is the horizontal distance from point A to point B? What is the vertical distance from point A to point B? horizontal distance: x 2 - x 1; vertical distance: y 2 - y 1 Module 10 GE_MNLESE385795_U3M10L3.indd 521 521 Lesson 3 02/04/14 5:46 AM DO NOT EDIT--Changes must be made through "File info" CorrectionKey=NL-A;CA-A E F DO NOT Correcti The horizontal and vertical distances from A to M must be half these distances. x2 - x1 2 What is the horizontal distance from point A to point M? y2 - y1 2 What is the vertical distance from point A to point M? _ _ To find the coordinates of point M, add the distances from Step E to the x- and y-coordinates of point A and simplify. y B (x2, y2) y2 M y1 x x1 A (x , y ) x2 1 1 2x x2 - x1 2x 1 + x 2 - x 1 _ x + x2 x2 - x1 _ x-coordinate of point M: x 1 + _ = 1+_ = __ = 1 2 2 2 2 2 2y 1 y2 - y1 2y 1 + y 2 - y 1 y1 + y2 y2 - y1 = + = = y1 + 2 2 2 2 2 y-coordinate of point M: _ _ _ __ _ Reflect 1. In the proof of the Distance Formula, why do you assume that x 1 ≠ x 2 and y 1 ≠ y 2? If x 1 = x 2 or y 1 = y 2, the segment is vertical or horizontal and it is not possible to form a right triangle as in the figure. 2. Does the Distance Formula still apply if x 1 = x 2 or y 1 = y 2? Explain. Yes; if x 1 = x 2, the formula simplifies to JK = ⎜y 2 - y 1⎟, and if y 1 = y 2, the formula simplifies to JK = ⎜x 2 - x 1⎟. 3. Does the Midpoint Formula still apply if x 1 = x 2 or y 1 = y 2? Explain. y1 + y2 , and if y 1 = y 2, the formula simplifies Yes; if x 1 = x 2, the formula simplifies to M x 1, 2 x1 + x2 , y1 . to M 2 (_ ) ( _) © Houghton Mifflin Harcourt Publishing Company Module 10 GE_MNLESE385795_U3M10L3.indd 522 522 Lesson 3 3/22/14 10:04 AM DO NOT EDIT--Changes must be made through "File info" CorrectionKey=NL-A;CA-A Explain 1 Positioning a Triangle on the Coordinate Plane A coordinate proof is a style of proof that uses coordinate geometry and algebra. The first step of a coordinate proof is to position the given figure in the plane. You can use any position, but some strategies can make the steps of the proof simpler. Strategies for Positioning Figures in the Coordinate Plane • Use the origin as a vertex, keeping the figure in Quadrant I. • Center the figure at the origin. • Center a side of the figure at the origin. • Use one or both axes as sides of the figure. Example 1 Write each coordinate proof. _ Given: ∠B is a right angle in △ABC. D is the midpoint of AC. Prove: The area of △DBC is one half the area of △ABC. Step 1 Assign coordinates to each vertex. Since you will use the Midpoint Formula to find the coordinates of D, use multiples of 2 for the leg lengths. The coordinates of A are (0, 2j). The coordinates of B are (0, 0). The coordinates of C are (2n, 0). Step 2 Position the figure on the coordinate plane. A (0, 2j) y Step 3 Write a coordinate proof. D △ABC is a right triangle with height 2j and base 2n. © Houghton Mifflin Harcourt Publishing Company area of △ABC = _12bh = _12(2n)(2j) = 2nj square units x B (0, 0) ( C (2n, 0) ) 2j + 0 0 + 2n _ , By the Midpoint Formula, the coordinates of D = _ = (n, j). 2 2 The height of △DBC is j units, and the base is 2n units. area of △DBC = _12bh = _12(2n)(j) = nj square units Since nj = _12 (2nj), the area of △DBC is one half the area of △ABC. Module10 GE_MNLESE385795_U3M10L3.indd 523 523 Lesson3 3/22/14 10:04 AM DO NOT EDIT--Changes must be made through "File info" CorrectionKey=NL-A;CA-A B DO NOT Correcti _ Given: ∠B is a right angle in △ABC. D is the midpoint of AC. Prove: The area of △ADB is one half the area of △ABC. Assign coordinates and position the figure as in Example 1A. △ABC is a right triangle with height 2j and base 2n . D x area of △ABC = _12bh = _12 2n ⋅ 2j = 2nj square units y A (0, 2j) B (0, 0) ( ) ( 0 + 2n 2j + 0 _______ = By the Midpoint Formula, the coordinates of D = _______ , 2 2 The height of △ADB is n units, and the base is 2j units. n , C (2n, 0) j ). area of △ADB = _12bh = _12 2j ⋅ n = jn square units _1 Since jn = 2 (2 nj) , the area of △ADB is one half the area of △ABC. Reflect 4. Why is it possible to position △ABC so that two of its sides lie on the axes of the coordinate plane? It is given that △ABC has a right angle at ∠B. The axes of the coordinate plane intersect at a right angle, so if vertex B is at the origin, then two sides of the triangle will lie on the axes. Your Turn Position the given triangle on the coordinate plane. Then show that the result about areas from Example 1 holds for the triangle. 5. 1 (2)(4) = 4 units 2 The height of △ABC is 4 units, and the base is 2 units. area of △ABC = __ 2 ¯, are The coordinates of D, the midpoint of AC +2 4 +0 , ____ = (1, 2). (0____ 2 2 ) 1 (2)(2) = 2 units 2 The height of △DBC is 2 units, and the base is 2 units. area of △DBC = __ 2 1 (4), the area of △DBC is one half the area of △ABC. Since 2 = __ 2 Module 10 GE_MNLESE385795_U3M10L3.indd 524 524 © Houghton Mifflin Harcourt Publishing Company A right triangle, △ABC, with legs of length 2 units and 4 units Possible answer: Let the coordinates of the vertices be A(0, 4), B(0, 0), and C(2, 0). Lesson 3 3/22/14 10:04 AM DO NOT EDIT--Changes must be made through "File info" CorrectionKey=NL-A;CA-A 6. A right triangle, △ABC, with both legs of length 8 units Possible answer: Let the coordinates of the vertices be A(0, 8), B(0, 0), and C(8, 0). 1 (8)(8) = 32units 2 The height of △ABC is 8 units, and the base is 8 units. area of △ABC = __ 2 ( ) +8 8 +0 ¯, are 0____ The coordinates of D, the midpoint of AC , ____ = (4, 4). 2 2 1 ( )( ) The height of △DBC is 4 units, and the base is 8 units. area of △DBC = _ 8 4 = 16 units2 2 1( ) Since 16 = _ 32 , the area of△DBC is one half the area of △ABC. 2 Explain 2 Proving the Triangle Midsegment Theorem In Module 8, you learned that the Triangle Midsegment Theorem states that a midsegment of a triangle is parallel to the third side of the triangle and is half as long as the third side. You can now use a coordinate proof to show that the theorem is true. Example 2 Prove the Triangle Midsegment Theorem. _ Given: XY is a midsegment of △PQR. _ _ Prove: XY ∥ PQ and XY = _12PQ P X Place △PQR so that one vertex is at the origin. For convenience, assign vertex P the coordinates (2a, 2b) and assign vertex Q the vertices (2c, 2d). ) y P (2a, 2b) 0 + 2b 0 + 2a _ , The coordinates of X are X _ = X( a,b ). 2 2 ( 0 + 0 2c + 2d ) Y R Use the Midpoint Formula to find the coordinates of X and Y. ( ( X ) Q (2c, 2d) , ___________ =Y c , d . The coordinates of Y are Y ___________ 2 2 R (0, 0) _ _ Find the slope of PQand XY. d - b d - b _ y2 - y1 _ _ y2 - y1 2d - 2b = ___________ ; slope of XY _ ___________ = = slope of PQ = _ = x2 - x1 x2 - x1 2c - 2a © Houghton Mifflin Harcourt Publishing Company c - Q a c - Y x a _ _ Therefore, PQ ǁ XY sincethe slopes are the same. Use the Distance Formula to find PQ and XY. ____ PQ = √(x 2 - x 1) 2 + ( y 2 - y 1 ) 2 = = ―――――――――――― √ 4 ⋅ (c - a) 2 + 4 ⋅ ( d - b ) 2 ―― √ ___ 4 ⋅ √(c - a) 2 + (d - b) 2 ____ XY = √ (x 2 - x 1) 2 + (y 2 - y 1) 2 ____ = √(2c - 2a) + (2d - 2b) 2 = = = 2 ――――――――― √4 ⋅ (c - a) 2 + (d - b) 2 ___ 2 √(c - a) 2 + (d - b) 2 _____ √( ) ( 2 c - a + d - b ) 2 1 This shows that XY = _____ PQ. 2 Module10 GE_MNLESE385795_U3M10L3.indd 525 525 Lesson3 3/22/14 10:04 AM DO NOT EDIT--Changes must be made through "File info" CorrectionKey=NL-A;CA-A DO NOT Correcti Reflect 7. Discussion Why is it more convenient to assign vertex P the coordinates (2a, 2b) and vertex Q the coordinates (2c, 2d) rather than using the coordinates (a, b) and (c, d)? ¯. Assigning ¯ and RQ The proof requires finding the coordinates of the midpoints of RP coordinates that are multiples of 2 eliminates the need for fractions in the coordinates of the midpoints. Explain 3 Proving the Concurrency of Medians Theorem You used the Concurrency of Medians Theorem in Module 8 and proved it in Module 9. Now you will prove the theorem again, this time using coordinate methods. Example 3 Prove the Concurrency of Medians Theorem. Q N _ _ _ Given: △PQR with medians PL, QM, and RN _ _ _ Prove: PL, QM, and RN are concurrent. P L M R Place △PQR so that vertex R is at the origin. Also, place the triangle so that point N lies on the y-axis. For convenience, assign point N the vertices (0, 6a). (The factor of 6 will result in easier calculations later.) y Q N (0, 6a) P x _ Since N is the midpoint of PQ, assign coordinates to P and Q as follows. y The horizontal distance from N to P must be the same as the horizontal distance from N to Q. Let this distance be 2b. Then the x-coordinate of point P is -2b and the x-coordinate of point Q is 2b . Q ( 2b , 6a + 2c ) N (0, 6a) P ( -2b , 6a - 2c ) The vertical distance from N to P must be the same as the vertical distance from N to Q. Let this distance be 2c. Then the y-coordinate of point P is 6a - 2c and the y-coordinate of point Q is 6a + 2c . © Houghton Mifflin Harcourt Publishing Company R (0, 0) x R (0, 0) Complete the figure by writing the coordinates of points P and Q. Module10 GE_MNLESE385795_U3M10L3.indd 526 526 Lesson3 3/22/14 10:04 AM DO NOT EDIT--Changes must be made through "File info" CorrectionKey=NL-A;CA-A Now use the Midpoint Formula to find the coordinates of L and M. ( ( ) (b 0 + 6a + 2c 0 + 2b _ The midpoint of RQ is L __ , __ = L 2 2 ) , 3a + c ( ) . 0 + -2b 0 + 6a - 2c _ 3a - c The midpoint of RP is M __ , __ = M -b , 2 2 ) . y Complete the figure by writing the coordinates of points L and M. ‹ › − To complete the proof, write the equation of QM and use the equation to find the coordinates of _ _point C, which is the intersection of the medians QM and RN. Then show that ‹ › − point C lies on PL . Q ( 2b , 6a + 2c ) N (0, 6a) P ( -2b , 6a - 2c ) C L ( b , 3a + c ) M ( -b , 3a - c ) ‹ › − Write the equation of QM using point-slope form. x R (0, 0) a + c 3 a +3 c ‹ › ( 6a + 2c ) - ( 3a - c ) − The slope of QM is __ = __ = __. 2b - (-b) 3 b b ‹ › − Use the coordinates of point Q for the point on QM . ) ( a + c ‹ › − Therefore, the equation of QM is y - 6a + 2c = __ ⋅ x - 2b . b Since point C lies on the y-axis, the x-coordinate of point C is 0. To find the y-coordinate of C, ‹ › − substitute x = 0 in the equation of QM and solve for y. a + c Substitute x = 0. y - 6a + 2c = __ ⋅ 0 - 2b ( © Houghton Mifflin Harcourt Publishing Company b Simplify the right side of the equation. y - 6a + 2c = -2 Distributive property y - 6a + 2c = -2 a - 2 c Add 6a + 2c to each side and simplify. (0 So, the coordinates of point C are C ) ) a+c y = 4a , 4a . ‹ › − Now write the equation of PL using point-slope form. 3 a -3 c a - c ‹ › ( 6a - 2c ) - (3a + c) − The slope of PL is __ = __ = __. -2b - b -3 b - b ‹ › − Use the coordinates of point P for the point on PL . ( ) a - c ‹ › − Therefore, the equation of PL is y - 6a - 2c = __ ⋅ x + 2b . - b Module 10 GE_MNLESE385795_U3M10L3.indd 527 527 Lesson 3 3/22/14 10:04 AM DO NOT EDIT--Changes must be made through "File info" CorrectionKey=NL-A;CA-A DO NOT Correcti ‹ › − Finally, show that point C lies on PL . To do so, show that when x = 0 in the ‹ › − equationfor PL , y = 4a. ( Substitute x = 0. a - c y - 6a - 2c = __ ⋅ 0 + 2b - b Simplify right side of equation. y - 6a - 2c = -2 a + 2 c Add 6a - 2c to each side and simplify. ) y = 4a Reflect 8. A student claims that the averages of the x-coordinates and of the y-coordinates of the vertices of the triangle are x- and y-coordinates of the point of concurrency, C. Does the coordinate proof of the Concurrency of Medians Theorem support the claim? Explain. Yes. The x-coordinates of the vertices of the triangle are −2b, 0, and 2b. The average of these coordinates is 0. The y-coordinates of the vertices of the triangle are 6a - 2c, 0, and 6a + 2c. The average of these coordinates is 4a. These averages match the coordinates of point C, (0, 4a). Using Triangles on the Coordinate Plane Explain 4 Example 4 Write each proof. Given: A(2, 3), B(5, −1), C(1, 0), D(-4, −1), E(0, 2), F(−1, −2) Prove: ∠ABC ≅ ∠DEF Step 1 Plot the points on a coordinate plane. y A E x ___ _ D ___ -2 0 C F 2 B _ 2 2 2 2 AB = √(5 - 2) + (-1 - 3) = √ 25 = 5; BC = √ (1 - 5) + 0 - (-1) = √17; ___ AC = √ (1 - 2) + (0 - 3) = 2 2 _ √ 10; ____ ____ DE = √(0 - (-4)) + (2 - (-1)) = √25 = 5; _ 2 2 _____ EF = √(-1 - 0) + (-2 - 2) = √1 +16 = √17; DF = √(-1 - (-4)) + (-2 - (-1)) 2 _ 2 _ _ 2 2 _ = √9 +1 = √10 _ _ _ _ _ _ So, AB ≅ DE, BC ≅ EF, and AC ≅ DF. Therefore, △ABC ≅ △DEF by the SSS Triangle Congruence Theorem and ∠ABC ≅ ∠DEF by CPCTC. Module10 GE_MNLESE385795_U3M10L3.indd 528 528 © Houghton Mifflin Harcourt Publishing Company Step 2 Use the Distance Formula to find the length of each side of each triangle. Lesson3 3/22/14 10:04 AM DO NOT EDIT--Changes must be made through "File info" CorrectionKey=NL-A;CA-A B _ Given: J(-4, 1), K(0, 5), L(3, 1), M(−1, −3), R is the midpoint of JK, S is the midpoint _ of LM. Prove: ∠JSK ≅ ∠LRM Step 1 Plot the points on a coordinate plane. y K 4 R ( ( ) ) ( ) 1 + 5 -4 + 0 R __ , __ = R -2 3 , 2 2 ( -4 ) 1 + -3 3 + -1 S __ , __ = S 1 -1 2 2 , Step 3 Use the Distance Formula to find the length of each side of each triangle. ____ JK = √(0 - (-4)) + (5 - 1) = √16 + 16 = √32 2 __ 2 ____ KS = JS = √( 1 √( 1 ) ( ) =√ 1 2 - 0 + -1 - 5 ) ( _____ - (-4) + -1 - 1 ____ 2 __ = __ √( -2 2 ) (3 ) _____ MR = © Houghton Mifflin Harcourt Publishing Company LR = 2 - (-1) + - (-3) ____ √( -2 - 3) + ( 3 2 2 S 2 4 M -4 4 _ = √ 29 _ __ = √ ) = √ 25 + -1 2 -2 √ 37 LM = √(-1 - 3) + (-3 - 1) = √16 + 16 = √32 2 0 x _ + 36 = ) √ 25 + 2 -2 L _ __ 2 2 J Step 2 Use the Midpoint Formula to find the coordinates of R and S. 1 + 36 = __ 4 = _ _ ¯ ¯ _ ¯ So, JK ≅ LM , KS ≅ MR , and JS ≅ LR . Therefore, △JKS ≅ _ _ √ 37 √ 29 △LMR by the SSS Triangle Congruence Theorem and ∠JSK ≅ ∠LRM corresponding parts of congruent triangles are congruent . since Reflect 9. In Part B, what other pairs of angles can you prove to be congruent? Why? ∠JKS ≅ ∠LMR and ∠KJS ≅ ∠MLR because these are corresponding parts of congruent triangles. Module 10 GE_MNLESE385795_U3M10L3.indd 529 529 Lesson 3 3/22/14 10:04 AM DO NOT EDIT--Changes must be made through "File info" CorrectionKey=NL-A;CA-A DO NOT Correcti Your Turn Write each proof. 10. Given: A(-4, −2), B(−3, 2), C(−1, 3), D(-5, 0), E(−1, −1), F(0, −3) Prove: ∠BCA ≅ ∠EFD ― ― ― ― ― By the Distance Formula, AB = √17 , BC = √5 , AC = √34 , DE = √17 , EF = √5 , ― ¯ ≅ EF ¯ ≅ DF ¯ ≅ DE ¯, BC ¯. Therefore, △ABC ≅ △DEF by the ¯, and AC DF = √34 . So, AB SSS Triangle Congruence Theorem and ∠BCA ≅ ∠EFD since corresponding parts of congruent triangles are congruent. _ _ 11. Given: P(−3, 5), Q(−1, −1), R(4, 5), S(2, −1), M is the midpoint of PQ, N is the midpoint of RS. Prove: ∠PQN ≅ ∠RSM By the Midpoint Formula, the coordinates of M are M the coordinates of N are N (____, _______) = N(3,2). 4 + 2 5 + (-1) 2 2 ― (________, _______) = M(-2, 2) and -3 + (-1) 5 + (-1) 2 2 ― ― ― By the Distance Formula, PQ = √40 , QN = 5, PN = √45 , RS = √40 , SM = 5, RM = √45 . ¯ ≅ RS ¯ ≅ SM ¯, QN ¯, and PN ¯ ≅ RM ¯. Therefore, △PQN ≅ △RSM by the SSS Triangle So, PQ Congruence Theorem and ∠PQN ≅ ∠RSM since corresponding parts of congruent triangles are congruent. Elaborate 12. When you write a coordinate proof, why might you assign 2p as a coordinate rather than p? If the proof includes finding the midpoint of a segment, assigning the coordinate 2p might make it possible to avoid fractions in the coordinates of the midpoint. geometry and algebra to draw conclusions. Coordinate proofs often involve the Distance Formula, the Midpoint Formula, and/or the Slope Formula. Module 10 GE_MNLESE385795_U3M10L3.indd 530 530 © Houghton Mifflin Harcourt Publishing Company 13. Essential Question Check-In What makes a coordinate proof different from the other types of proofs you have written so far? A coordinate proof depends upon assigning coordinates to a figure and using coordinate Lesson 3 3/22/14 10:04 AM