Coordinate Geometry Proofs

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Answer Key & Explanations

Coordinate Geometry Proofs · Grade 10 · Worksheet 1

1. Prove that quadrilateral ABCD with vertices A(8, 10), B(14, 18), C(20, 10), and D(14, 2) is a rhombus using coordinate geometry. Answer: Quadrilateral ABCD is a rhombus. Solution: Use the distance formula d = sqrt((x2 - x1)^2 + (y2 - y1)^2) to find the lengths of all four sides.
   Full step-by-step solution

   Step 1: Use the distance formula d = sqrt((x2 - x1)^2 + (y2 - y1)^2) to find the lengths of all four sides. Step 2: Calculate AB: A(8, 10), B(14, 18) d_AB = sqrt((14 - 8)^2 + (18 - 10)^2) = sqrt(6^2 + 8^2) = sqrt(36 + 64) = sqrt(100) = 10 Step 3: Calculate BC: B(14, 18), C(20, 10) d_BC = sqrt((20 - 14)^2 + (10 - 18)^2) = sqrt(6^2 + (-8)^2) = sqrt(36 + 64) = sqrt(100) = 10 Step 4: Calculate CD: C(20, 10), D(14, 2) d_CD = sqrt((14 - 20)^2 + (2 - 10)^2) = sqrt((-6)^2 + (-8)^2) = sqrt(36 + 64) = sqrt(100) = 10 Step 5: Calculate DA: D(14, 2), A(8, 10) d_DA = sqrt((8 - 14)^2 + (10 - 2)^2) = sqrt((-6)^2 + 8^2) = sqrt(36 + 64) = sqrt(100) = 10 Step 6: Since AB = BC = CD = DA = 10, all four sides are equal. A quadrilateral with all sides equal is a rhombus. The answer is: Quadrilateral ABCD is a rhombus.

2. Mason is analyzing a quadrilateral with vertices at P(8, 12), Q(18, 10), R(14, 0), and S(4, 2). Using the slope formula, prove whether PQRS is a parallelogram. Answer: Yes, PQRS is a parallelogram because opposite sides have equal slopes (PQ and SR both have slope -1/5; QR and SP both have slope -5/2). Solution: Identify the vertices in order: P(8,12), Q(18,10), R(14,0), S(4,2). Calculate slope of PQ: (10 - 12) / (18 - 8) = (-2) / 10 = -1/5. Calculate slope of QR: (0 - 10) / (14 - 18) = (-10) / (-4) = 5/2.
   Full step-by-step solution

   Step 1: Identify the vertices in order: P(8,12), Q(18,10), R(14,0), S(4,2). Step 2: Calculate slope of PQ: (10 - 12) / (18 - 8) = (-2) / 10 = -1/5. Step 3: Calculate slope of QR: (0 - 10) / (14 - 18) = (-10) / (-4) = 5/2. Step 4: Calculate slope of RS: (2 - 0) / (4 - 14) = 2 / (-10) = -1/5. Step 5: Calculate slope of SP: (12 - 2) / (8 - 4) = 10 / 4 = 5/2. Step 6: Compare opposite sides: PQ and RS both have slope -1/5, so they are parallel. QR and SP both have slope 5/2, so they are parallel. Step 7: Since both pairs of opposite sides are parallel, quadrilateral PQRS is a parallelogram. The answer is: Yes, PQRS is a parallelogram because opposite sides have equal slopes (PQ and SR both have slope -1/5; QR and SP both have slope -5/2).

3. A circle is centered at point (2, -1) on the coordinate plane and passes through point (5, 3). A tangent line is drawn to the circle at point (5, 3). Determine the equation of this tangent line in slope-intercept form. Answer: y = -3/4x + 27/4 Solution: Find the radius slope from center to point (5,3). The center is (2, -1). Slope of radius = (3 - (-1)) / (5 - 2) = (3 + 1) / 3 = 4 / 3.
   Full step-by-step solution

   Step 1: Find the radius slope from center to point (5,3). The center is (2, -1). Slope of radius = (3 - (-1)) / (5 - 2) = (3 + 1) / 3 = 4 / 3. Step 2: Tangent slope is perpendicular to radius. Perpendicular slope = negative reciprocal of 4/3 = -3/4. Step 3: Use point-slope form at point (5,3). Point-slope form: y - 3 = (-3/4)(x - 5). Step 4: Simplify to slope-intercept form. y - 3 = (-3/4)x + (15/4) y = (-3/4)x + 15/4 + 3 y = (-3/4)x + 15/4 + 12/4 y = (-3/4)x + 27/4. Final answer: y = -3/4x + 27/4.

4. Liam is designing a triangular garden with vertices at coordinates A(2, 3), B(8, 7), and C(11, 1) on a coordinate grid. He needs to prove that the garden forms a right triangle to ensure proper irrigation system placement. Using coordinate geometry, determine which vertex contains the right angle and calculate the area of the triangular garden in square units. Answer: Vertex B, area = 20 square units Solution: In coordinate geometry, a triangle is right-angled if two of its sides are perpendicular, which occurs when the product of their slopes equals -1.
   Full step-by-step solution

   In coordinate geometry, a triangle is right-angled if two of its sides are perpendicular, which occurs when the product of their slopes equals -1. The area of a right triangle can be found by taking half the product of the lengths of the two legs (the sides forming the right angle). This method combines slope analysis with distance calculations to solve geometric problems.

5. Liam is analyzing a quadrilateral on the coordinate plane with vertices at A(-3, 1), B(3, -5), C(9, 1), and D(3, 7). Using coordinate geometry, prove that quadrilateral ABCD is a rhombus. Answer: Quadrilateral ABCD is a rhombus because all four sides are equal in length (each side = sqrt(72) = 6*sqrt(2) units) and the diagonals are perpendicular (slope of AC = 0, slope of BD is undefined, so they are perpendicular). Solution: Label vertices in order: A(-3, 1), B(3, -5), C(9, 1), D(3, 7). Compute side lengths using distance formula d = sqrt((x2-x1)^2 + (y2-y1)^2).
   Full step-by-step solution

   Step 1: Label vertices in order: A(-3, 1), B(3, -5), C(9, 1), D(3, 7). Step 2: Compute side lengths using distance formula d = sqrt((x2-x1)^2 + (y2-y1)^2). Side AB: sqrt((3 - (-3))^2 + (-5 - 1)^2) = sqrt(6^2 + (-6)^2) = sqrt(36 + 36) = sqrt(72). Side BC: sqrt((9 - 3)^2 + (1 - (-5))^2) = sqrt(6^2 + 6^2) = sqrt(36 + 36) = sqrt(72). Side CD: sqrt((3 - 9)^2 + (7 - 1)^2) = sqrt((-6)^2 + 6^2) = sqrt(36 + 36) = sqrt(72). Side DA: sqrt((-3 - 3)^2 + (1 - 7)^2) = sqrt((-6)^2 + (-6)^2) = sqrt(36 + 36) = sqrt(72). Step 3: All four sides equal sqrt(72) = 6*sqrt(2) units, so ABCD is a rhombus. Step 4: (Optional) Check diagonals: AC from A(-3,1) to C(9,1): slope = (1-1)/(9-(-3)) = 0/12 = 0 (horizontal). BD from B(3,-5) to D(3,7): slope = (7-(-5))/(3-3) = 12/0, which is undefined (vertical). Horizontal and vertical lines are perpendicular, confirming a property of a rhombus. Conclusion: Quadrilateral ABCD is a rhombus.

6. A circle is centered at the origin with radius 5 units. A line with equation y = 2x + 3 intersects this circle. Find the coordinates of the intersection points by solving the system of equations algebraically. Answer: (-2, -1) and (1.6, 6.2) Solution: In coordinate geometry, finding intersection points between a line and a circle involves solving a system of equations.
   Full step-by-step solution

   In coordinate geometry, finding intersection points between a line and a circle involves solving a system of equations. The circle equation comes from the distance formula, while the line equation provides a relationship between x and y. Substituting the linear expression into the circle equation yields a quadratic, whose solutions correspond to the intersection points. This method works for any line and circle configuration.

7. Prove that quadrilateral PQRS with vertices P(2, 7), Q(7, 12), R(12, 7), and S(7, 2) is a rhombus using coordinate geometry. Answer: PQRS is a rhombus because all four sides are equal in length (PQ = QR = RS = SP = 5√2). Solution: Find PQ: P(2,7), Q(7,12). PQ = sqrt((7-2)^2 + (12-7)^2) = sqrt(5^2 + 5^2) = sqrt(25+25) = sqrt(50) = 5√2. Find QR: Q(7,12), R(12,7).
   Full step-by-step solution

   Step 1: Apply the distance formula: d = sqrt((x2 - x1)^2 + (y2 - y1)^2). Step 2: Find PQ: P(2,7), Q(7,12). PQ = sqrt((7-2)^2 + (12-7)^2) = sqrt(5^2 + 5^2) = sqrt(25+25) = sqrt(50) = 5√2. Step 3: Find QR: Q(7,12), R(12,7). QR = sqrt((12-7)^2 + (7-12)^2) = sqrt(5^2 + (-5)^2) = sqrt(25+25) = sqrt(50) = 5√2. Step 4: Find RS: R(12,7), S(7,2). RS = sqrt((7-12)^2 + (2-7)^2) = sqrt((-5)^2 + (-5)^2) = sqrt(25+25) = sqrt(50) = 5√2. Step 5: Find SP: S(7,2), P(2,7). SP = sqrt((2-7)^2 + (7-2)^2) = sqrt((-5)^2 + 5^2) = sqrt(25+25) = sqrt(50) = 5√2. Step 6: Since PQ = QR = RS = SP = 5√2, all four sides are equal. Therefore, quadrilateral PQRS is a rhombus.