Coordinate Geometry Proofs

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Coordinate Geometry Proofs · Grade 10 · Worksheet 2

1. Mason is a land surveyor mapping a new housing development. He marks four points on a coordinate grid to represent the corners of a proposed park: A(-7, 2), B(2, 7), C(7, -2), and D(-2, -7). Mason needs to prove that the quadrilateral formed by these points is a square by using coordinate geometry. Determine whether ABCD is a square by calculating the side lengths, the slopes of adjacent sides, and the lengths of the diagonals. Answer: Yes, ABCD is a square. All sides have length sqrt(106), adjacent sides are perpendicular (slopes are negative reciprocals: 5/9 and -9/5), and diagonals have equal length sqrt(212). Solution: Calculate side lengths using distance formula. Distance AB = sqrt((2 - (-7))^2 + (7 - 2)^2) = sqrt(9^2 + 5^2) = sqrt(81 + 25) = sqrt(106) Distance BC = sqrt((7 - 2)^2 + (-2 - 7)^2) = sqrt(5^2 + (-9)^2) = sqrt(25 + 81) = sqrt(106) Distance CD = sqrt((-2 - 7)^2 + (-7 - (-2))^2) = sqrt((-9)^2 +…
   Full step-by-step solution

   Step 1: Calculate side lengths using distance formula. Distance AB = sqrt((2 - (-7))^2 + (7 - 2)^2) = sqrt(9^2 + 5^2) = sqrt(81 + 25) = sqrt(106) Distance BC = sqrt((7 - 2)^2 + (-2 - 7)^2) = sqrt(5^2 + (-9)^2) = sqrt(25 + 81) = sqrt(106) Distance CD = sqrt((-2 - 7)^2 + (-7 - (-2))^2) = sqrt((-9)^2 + (-5)^2) = sqrt(81 + 25) = sqrt(106) Distance DA = sqrt((-7 - (-2))^2 + (2 - (-7))^2) = sqrt((-5)^2 + 9^2) = sqrt(25 + 81) = sqrt(106) All four sides have length sqrt(106). Step 2: Check slopes of adjacent sides. Slope AB = (7 - 2) / (2 - (-7)) = 5/9 Slope BC = (-2 - 7) / (7 - 2) = -9/5 Since (5/9) * (-9/5) = -1, AB is perpendicular to BC. Similarly, slope CD = (-7 - (-2)) / (-2 - 7) = -5 / -9 = 5/9 Slope DA = (2 - (-7)) / (-7 - (-2)) = 9 / -5 = -9/5 So adjacent sides are perpendicular. Step 3: Calculate diagonal lengths. Diagonal AC = sqrt((7 - (-7))^2 + (-2 - 2)^2) = sqrt(14^2 + (-4)^2) = sqrt(196 + 16) = sqrt(212) Diagonal BD = sqrt((-2 - 2)^2 + (-7 - 7)^2) = sqrt((-4)^2 + (-14)^2) = sqrt(16 + 196) = sqrt(212) Both diagonals have equal length sqrt(212). Since all sides are equal, adjacent sides are perpendicular, and diagonals are equal, quadrilateral ABCD is a square.

2. Charlotte is analyzing a quadrilateral on a coordinate plane with vertices at A(8, 12), B(18, 10), C(14, 0), and D(4, 2). Using the slope formula, prove whether quadrilateral ABCD is a parallelogram. Answer: Yes, quadrilateral ABCD is a parallelogram because opposite sides have equal slopes: AB and CD both have slope -1/5, and BC and DA both have slope -5/2. Solution: Calculate the slope of AB. Coordinates: A(8, 12), B(18, 10). Slope = (10 - 12)/(18 - 8) = (-2)/10 = -1/5.
   Full step-by-step solution

   Step 1: Calculate the slope of AB. Coordinates: A(8, 12), B(18, 10). Slope = (10 - 12)/(18 - 8) = (-2)/10 = -1/5. Step 2: Calculate the slope of BC. Coordinates: B(18, 10), C(14, 0). Slope = (0 - 10)/(14 - 18) = (-10)/(-4) = 5/2. Step 3: Calculate the slope of CD. Coordinates: C(14, 0), D(4, 2). Slope = (2 - 0)/(4 - 14) = 2/(-10) = -1/5. Step 4: Calculate the slope of DA. Coordinates: D(4, 2), A(8, 12). Slope = (12 - 2)/(8 - 4) = 10/4 = 5/2. Step 5: Compare slopes: AB and CD both have slope -1/5, so AB is parallel to CD. BC and DA both have slope 5/2, so BC is parallel to DA. Since both pairs of opposite sides are parallel, quadrilateral ABCD is a parallelogram.

3. Isabella is analyzing a quadrilateral on a coordinate plane with vertices at W(10, 9), X(16, 15), Y(10, 21), and Z(4, 15). Prove that this quadrilateral is a rhombus using the distance formula. Then, prove that it is also a square using the slope formula. Answer: The quadrilateral is a rhombus because all four sides have length sqrt(72) = 6*sqrt(2). It is also a square because adjacent sides are perpendicular (slopes of WX and XY are 1 and -1, respectively, and their product is -1). Solution: Calculate the length of each side using the distance formula: d = sqrt((x2 - x1)^2 + (y2 - y1)^2).
   Full step-by-step solution

   Step 1: Calculate the length of each side using the distance formula: d = sqrt((x2 - x1)^2 + (y2 - y1)^2). Side WX: W(10,9), X(16,15) WX = sqrt((16-10)^2 + (15-9)^2) = sqrt(6^2 + 6^2) = sqrt(36 + 36) = sqrt(72) = 6*sqrt(2) Side XY: X(16,15), Y(10,21) XY = sqrt((10-16)^2 + (21-15)^2) = sqrt((-6)^2 + 6^2) = sqrt(36 + 36) = sqrt(72) = 6*sqrt(2) Side YZ: Y(10,21), Z(4,15) YZ = sqrt((4-10)^2 + (15-21)^2) = sqrt((-6)^2 + (-6)^2) = sqrt(36 + 36) = sqrt(72) = 6*sqrt(2) Side ZW: Z(4,15), W(10,9) ZW = sqrt((10-4)^2 + (9-15)^2) = sqrt(6^2 + (-6)^2) = sqrt(36 + 36) = sqrt(72) = 6*sqrt(2) All four sides have length sqrt(72) = 6*sqrt(2), so the quadrilateral is a rhombus. Step 2: Calculate the slopes of adjacent sides to check for perpendicularity. Slope formula: m = (y2 - y1)/(x2 - x1). Slope of WX: (15-9)/(16-10) = 6/6 = 1 Slope of XY: (21-15)/(10-16) = 6/(-6) = -1 The product of the slopes is 1 * (-1) = -1. Since the product of the slopes of adjacent sides is -1, side WX is perpendicular to side XY. Step 3: Since the quadrilateral is a rhombus with one pair of adjacent sides perpendicular, all interior angles are right angles. Therefore, the quadrilateral is a square. The quadrilateral is a rhombus (all sides equal) and a square (all sides equal and all angles 90 degrees).

4. Prove that quadrilateral SOPHIA with vertices S(7, 9), O(15, 13), P(19, 21), H(11, 17) is a parallelogram using slopes. Answer: Quadrilateral SOPHIA is a parallelogram because opposite sides have equal slopes: slope(SO) = slope(PH) = 1/2 and slope(OP) = slope(HS) = 2. Solution: Find slope of SO: S(7,9), O(15,13). m_SO = (13-9)/(15-7) = 4/8 = 1/2. Find slope of OP: O(15,13), P(19,21).
   Full step-by-step solution

   Step 1: Find slope of SO: S(7,9), O(15,13). m_SO = (13-9)/(15-7) = 4/8 = 1/2. Step 2: Find slope of OP: O(15,13), P(19,21). m_OP = (21-13)/(19-15) = 8/4 = 2. Step 3: Find slope of PH: P(19,21), H(11,17). m_PH = (17-21)/(11-19) = (-4)/(-8) = 1/2. Step 4: Find slope of HS: H(11,17), S(7,9). m_HS = (9-17)/(7-11) = (-8)/(-4) = 2. Step 5: Compare slopes: m_SO = m_PH = 1/2, so SO is parallel to PH. m_OP = m_HS = 2, so OP is parallel to HS. Since both pairs of opposite sides are parallel, quadrilateral SOPHIA is a parallelogram.

5. Prove that quadrilateral ABCD with vertices A(2, 7), B(7, 12), C(12, 7), and D(7, 2) is a rhombus using the distance formula. Answer: Rhombus Solution: Side AB: A(2,7), B(7,12). d_AB = sqrt((7-2)^2 + (12-7)^2) = sqrt(5^2 + 5^2) = sqrt(25+25) = sqrt(50) = 5*sqrt(2). Side BC: B(7,12), C(12,7).
   Full step-by-step solution

   Step 1: Apply the distance formula d = sqrt((x2 - x1)^2 + (y2 - y1)^2) to each side. Step 2: Side AB: A(2,7), B(7,12). d_AB = sqrt((7-2)^2 + (12-7)^2) = sqrt(5^2 + 5^2) = sqrt(25+25) = sqrt(50) = 5*sqrt(2). Step 3: Side BC: B(7,12), C(12,7). d_BC = sqrt((12-7)^2 + (7-12)^2) = sqrt(5^2 + (-5)^2) = sqrt(25+25) = sqrt(50) = 5*sqrt(2). Step 4: Side CD: C(12,7), D(7,2). d_CD = sqrt((7-12)^2 + (2-7)^2) = sqrt((-5)^2 + (-5)^2) = sqrt(25+25) = sqrt(50) = 5*sqrt(2). Step 5: Side DA: D(7,2), A(2,7). d_DA = sqrt((2-7)^2 + (7-2)^2) = sqrt((-5)^2 + 5^2) = sqrt(25+25) = sqrt(50) = 5*sqrt(2). Step 6: All four sides are equal to 5*sqrt(2). Therefore, quadrilateral ABCD is a rhombus.

6. A parabola is defined by the equation y = x^2 - 6x + 8. A line with equation y = 2x - 4 intersects this parabola at two points, forming a chord. Calculate the exact length of this chord using the distance formula. Answer: 4√5 Solution: Find the intersection points by setting the equations equal: x^2 - 6x + 8 = 2x - 4 Rearrange to form a quadratic equation: x^2 - 8x + 12 = 0 Factor the quadratic: (x - 2)(x - 6) = 0 Solve for x: x = 2 and x = 6 Find the corresponding y-coordinates using y = 2x - 4: When x = 2: y = 2(2) - 4 = 0,…
   Full step-by-step solution

   Step 1: Find the intersection points by setting the equations equal: x^2 - 6x + 8 = 2x - 4 Step 2: Rearrange to form a quadratic equation: x^2 - 8x + 12 = 0 Step 3: Factor the quadratic: (x - 2)(x - 6) = 0 Step 4: Solve for x: x = 2 and x = 6 Step 5: Find the corresponding y-coordinates using y = 2x - 4: When x = 2: y = 2(2) - 4 = 0, so point A is (2, 0) When x = 6: y = 2(6) - 4 = 8, so point B is (6, 8) Step 6: Apply the distance formula: d = √[(6-2)^2 + (8-0)^2] = √[4^2 + 8^2] = √[16 + 64] = √80 = √(16×5) = 4√5 The exact length of the chord is 4√5.

7. Mason is designing a kite for a school project. He plots the vertices of the kite on a coordinate plane at points P(-2, 2), Q(2, 7), R(7, 2), and S(2, -3). Mason wants to prove that his quadrilateral is a kite by showing that the diagonals are perpendicular. Using coordinate geometry, calculate the slopes of diagonals PR and QS, and determine whether the diagonals are perpendicular. Answer: Yes, the diagonals are perpendicular because the product of their slopes is -1. Solution: Find the coordinates of the endpoints of diagonal PR: P(-2, 2) and R(7, 2). Calculate the slope of PR using the slope formula m = (y2 - y1) / (x2 - x1). m_PR = (2 - 2) / (7 - (-2)) = 0 / 9 = 0.
   Full step-by-step solution

   Step 1: Find the coordinates of the endpoints of diagonal PR: P(-2, 2) and R(7, 2). Step 2: Calculate the slope of PR using the slope formula m = (y2 - y1) / (x2 - x1). m_PR = (2 - 2) / (7 - (-2)) = 0 / 9 = 0. Step 3: Find the coordinates of the endpoints of diagonal QS: Q(2, 7) and S(2, -3). Step 4: Calculate the slope of QS. m_QS = (-3 - 7) / (2 - 2) = (-10) / 0, which is undefined (vertical line). Step 5: A horizontal line (slope 0) and a vertical line (undefined slope) are perpendicular because the product of their slopes is 0 * undefined, which is interpreted as -1 in coordinate geometry. Thus, diagonals PR and QS are perpendicular, proving the quadrilateral is a kite.

8. Prove that quadrilateral Aroha with vertices A(1, 3), R(5, 7), O(9, 3), and H(5, -1) is a rhombus using coordinate geometry. Answer: It is a rhombus because all four sides are equal in length (each side = 4√2 units) and the diagonals are perpendicular (slopes are 0 and undefined). Solution: Find the lengths of all sides using the distance formula d = sqrt((x2 - x1)^2 + (y2 - y1)^2).
   Full step-by-step solution

   Step 1: Find the lengths of all sides using the distance formula d = sqrt((x2 - x1)^2 + (y2 - y1)^2). Side AR: A(1,3) to R(5,7) d = sqrt((5-1)^2 + (7-3)^2) = sqrt(4^2 + 4^2) = sqrt(16+16) = sqrt(32) = 4√2 Side RO: R(5,7) to O(9,3) d = sqrt((9-5)^2 + (3-7)^2) = sqrt(4^2 + (-4)^2) = sqrt(16+16) = sqrt(32) = 4√2 Side OH: O(9,3) to H(5,-1) d = sqrt((5-9)^2 + (-1-3)^2) = sqrt((-4)^2 + (-4)^2) = sqrt(16+16) = sqrt(32) = 4√2 Side HA: H(5,-1) to A(1,3) d = sqrt((1-5)^2 + (3-(-1))^2) = sqrt((-4)^2 + 4^2) = sqrt(16+16) = sqrt(32) = 4√2 All four sides are equal (4√2 units), so it is a rhombus. Step 2: Check if diagonals are perpendicular using slope formula m = (y2 - y1)/(x2 - x1). Diagonal AO: A(1,3) to O(9,3) Slope = (3-3)/(9-1) = 0/8 = 0 (horizontal line) Diagonal RH: R(5,7) to H(5,-1) Slope = (-1-7)/(5-5) = -8/0 = undefined (vertical line) Since one diagonal is horizontal (slope 0) and the other is vertical (undefined slope), the diagonals are perpendicular. Therefore, quadrilateral Aroha is a rhombus because all sides are equal and the diagonals are perpendicular.