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Question 1 / 20
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Given a point with coordinates (k, 2k - 1), if its ordinate and abscissa are equal, determine the value of k.
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Solution: Step 1: Understand that the abscissa refers to the x-coordinate and the ordinate refers to the y-coordinate. Step 2: For the given point (k, 2k - 1), the abscissa is k and the ordinate is (2k - 1). Step 3: The problem states that the ordinate and abscissa are equal, so set up the equation: k = 2k - 1. Step 4: Solve the equation for k: 1 = 2k - k => k = 1. Step 5: The value of k is 1.
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The intersection point of the lines 2x + 3y = 11 and x - 2y + 12 = 0 is P(x1, y1). The line x - 2y + 12 = 0 intersects the x-axis at Q(x2, y2). Calculate the value of (x1 - x2 + y1 + y2).
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Solution: Step 1: Find the intersection point P(x1, y1) by solving the system of equations: (1) 2x + 3y = 11 (2) x - 2y = -12 Step 2: Multiply equation (2) by 2: (3) 2x - 4y = -24 Step 3: Subtract equation (3) from equation (1): (2x + 3y) - (2x - 4y) = 11 - (-24) 7y = 35 y = 5 Step 4: Substitute y = 5 into equation (1) to find x: 2x + 3*(5) = 11 2x + 15 = 11 2x = -4 x = -2 So, P(x1, y1) = (-2, 5). Step 5: Find the point Q(x2, y2) where the line x - 2y + 12 = 0 intersects the x-axis. On the x-axis, y = 0. x - 2*(0) + 12 = 0 x + 12 = 0 x = -12 So, Q(x2, y2) = (-12, 0). Step 6: Calculate the value of (x1 - x2 + y1 + y2). (-2) - (-12) + 5 + 0 = -2 + 12 + 5 + 0 = 10 + 5 = 15.
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A line originates from the origin and intersects the line 3x - 2y = 6 perpendicularly at point M. Determine the coordinates of point M.
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Solution: Step 1: Find the slope of the given line 3x - 2y = 6. Rearrange to slope-intercept form (y = mx + c): -2y = -3x + 6 => y = (3/2)x - 3. So, the slope of the given line (m1) = 3/2. Step 2: Find the slope of the line perpendicular to it. For perpendicular lines, m1 * m2 = -1. (3/2) * m2 = -1 => m2 = -2/3. Step 3: Write the equation of the line passing through the origin (0, 0) with slope m2 = -2/3. Using y - y1 = m(x - x1): y - 0 = (-2/3)(x - 0) => y = (-2/3)x => 3y = -2x => 2x + 3y = 0. Step 4: Find the intersection point M by solving the system of equations: 1) 3x - 2y = 6 2) 2x + 3y = 0 Step 5: From equation (2), x = -3y/2. Step 6: Substitute this x into equation (1): 3(-3y/2) - 2y = 6 -9y/2 - 2y = 6 Multiply by 2: -9y - 4y = 12 -13y = 12 y = -12/13 Step 7: Substitute y back into x = -3y/2: x = -3/2 * (-12/13) x = (3 * 6) / 13 = 18/13 Step 8: The coordinates of point M are (18/13, -12/13).
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Determine the area (in square units) of the region bounded by the lines defined by the equations 2x - 3y + 6 = 0, 4x + y = 16, and y = 0.
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Solution: Step 1: Find the intersection points of the given lines to determine the vertices of the enclosed region. Step 2: Intersection of 2x - 3y + 6 = 0 and y = 0: Substitute y=0 into 2x - 3y + 6 = 0 => 2x + 6 = 0 => 2x = -6 => x = -3. Vertex 1: (-3, 0). Step 3: Intersection of 4x + y = 16 and y = 0: Substitute y=0 into 4x + y = 16 => 4x = 16 => x = 4. Vertex 2: (4, 0). Step 4: Intersection of 2x - 3y + 6 = 0 and 4x + y = 16: From 4x + y = 16, y = 16 - 4x. Substitute into 2x - 3y + 6 = 0: 2x - 3(16 - 4x) + 6 = 0 => 2x - 48 + 12x + 6 = 0 => 14x - 42 = 0 => 14x = 42 => x = 3. Substitute x=3 back into y = 16 - 4x: y = 16 - 4(3) = 16 - 12 = 4. Vertex 3: (3, 4). Step 5: The vertices of the triangle are A(-3, 0), B(4, 0), and C(3, 4). Step 6: The base of the triangle lies on the x-axis (y=0) from x=-3 to x=4. Length of base = 4 - (-3) = 7 units. Step 7: The height of the triangle is the perpendicular distance from vertex C to the x-axis, which is the y-coordinate of C, i.e., 4 units. Step 8: Calculate the area of the triangle: Area = (1/2) * base * height = (1/2) * 7 * 4 = 14 square units.
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Given parallelogram ABCD with coordinates A(5, 0), B(-2, 3), and C(-1, 4), find the equation of line AD.
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Solution: Step 1: In a parallelogram, opposite sides are parallel. Therefore, line AD is parallel to line BC. Step 2: Calculate the slope (m_BC) of line BC using points B(-2, 3) and C(-1, 4). m_BC = (y2 - y1) / (x2 - x1) = (4 - 3) / (-1 - (-2)) = 1 / (-1 + 2) = 1 / 1 = 1. Step 3: Since AD is parallel to BC, the slope of line AD (m_AD) is equal to m_BC. m_AD = 1. Step 4: Use the point-slope form of a linear equation (y - y1 = m(x - x1)) with point A(5, 0) and slope m_AD = 1. y - 0 = 1*(x - 5). y = x - 5.
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Two individuals, A and B, start from the same point. A travels 20 km in a north-east direction, while B travels 16 km east and then 12 km north. What is the distance between A and B?
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Solution: Step 1: Analyze A's movement - A travels 20 km in a north-east direction. Step 2: Analyze B's movement - B travels 16 km east and then 12 km north. Step 3: Apply Pythagorean theorem - Calculate the distance using the Pythagorean theorem: $\sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20$ km. Step 4: Compare with A's distance - Since A also travels 20 km, and in a manner that aligns with B's final position, determine if they end up at the same place. Step 5: Resolve the actual distance - Recognize that A's path forms a right-angled triangle where both legs align with B's movements, indicating they end up at the same point. Step 6: Conclude the distance between A and B - The distance between A and B is 0 km.
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At which point do the graphs of the equations 2x + 1 = 0 and 3y - 9 = 0 intersect?
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Solution: Step 1: Solve the first equation for x: 2x + 1 = 0 2x = -1 x = -1/2 Step 2: Solve the second equation for y: 3y - 9 = 0 3y = 9 y = 9/3 y = 3 Step 3: The point of intersection is given by the (x, y) coordinates. Step 4: Therefore, the point of intersection is (-1/2, 3).
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A point Q(a, b) is reflected across the y-axis to obtain Q1. Then, Q1 is reflected across the x-axis, resulting in the point (-5, 3). Determine the coordinates of point Q.
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Solution: Step 1: Let the final reflected point be Q2(-5, 3). This point Q2 is the reflection of Q1 across the x-axis. Step 2: If a point (x, y) is reflected across the x-axis, its new coordinates are (x, -y). Working backward, if Q2(-5, 3) is the reflection of Q1 across the x-axis, then Q1 must be (-5, -3). Step 3: Now, Q1(-5, -3) is the reflection of the original point Q(a, b) across the y-axis. Step 4: If a point (x, y) is reflected across the y-axis, its new coordinates are (-x, y). Working backward, if Q1(-5, -3) is the reflection of Q across the y-axis, then Q must be (5, -3). Step 5: Therefore, the coordinates of point Q are (5, -3).
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Find the ratio in which the point T(x, 0) divides the line segment connecting points S(-4, -1) and U(1, 4).
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Solution: Step 1: Let the point T(x, 0) divide the segment joining S(-4, -1) and U(1, 4) in the ratio k : 1. Step 2: Use the section formula for the y-coordinate: y = (k * y2 + 1 * y1) / (k + 1) Step 3: Substitute the coordinates of S(x1, y1) = (-4, -1), U(x2, y2) = (1, 4), and T(x, y) = (x, 0). 0 = (k * 4 + 1 * (-1)) / (k + 1) Step 4: Simplify the equation: 0 = (4k - 1) / (k + 1) Step 5: Multiply both sides by (k + 1): 0 * (k + 1) = 4k - 1 0 = 4k - 1 Step 6: Solve for k: 4k = 1 k = 1/4. Step 7: The ratio k : 1 is 1/4 : 1, which simplifies to 1 : 4. Step 8: (Optional) We can also find x using the x-coordinate section formula: x = (k * x2 + 1 * x1) / (k + 1) = ((1/4)*1 + 1*(-4)) / (1/4+1) = (1/4 - 4) / (5/4) = (-15/4) / (5/4) = -15/5 = -3. So the point is T(-3, 0).
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The line 3x + 4y = 24 intersects the x-axis at point A and the y-axis at point B. P(2, 0) and Q(0, 3/2) are points on segments OA and OB, respectively, where O is the origin. If AB = 10 cm, what is the length of PQ?
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Solution: Step 1: Find the x-intercept (point A) by setting y = 0 in the equation 3x + 4y = 24. 3x + 4(0) = 24 => 3x = 24 => x = 8. So, A = (8, 0). Step 2: Find the y-intercept (point B) by setting x = 0 in the equation 3x + 4y = 24. 3(0) + 4y = 24 => 4y = 24 => y = 6. So, B = (0, 6). Step 3: Identify the given points P(2, 0) and Q(0, 3/2). (Note: The information AB = 10 cm is consistent but not strictly needed to find PQ. Distance AB = sqrt((8-0)^2 + (0-6)^2) = sqrt(64+36) = sqrt(100) = 10.) Step 4: Use the distance formula to find the length of PQ, where P=(x1, y1)=(2, 0) and Q=(x2, y2)=(0, 3/2). Distance PQ = sqrt((x2 - x1)^2 + (y2 - y1)^2) Step 5: Substitute the coordinates: PQ = sqrt((0 - 2)^2 + (3/2 - 0)^2) PQ = sqrt((-2)^2 + (3/2)^2) PQ = sqrt(4 + 9/4) PQ = sqrt(16/4 + 9/4) PQ = sqrt(25/4) PQ = 5/2 Step 6: Convert to decimal: PQ = 2.5 cm.
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Calculate the area of the quadrilateral defined by connecting the points (4, 2), (8, 2), (8, 14), and (4, 10).
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Solution: Step 1: Label the given points as A(4, 2), B(8, 2), C(8, 14), and D(4, 10). Step 2: Divide the quadrilateral ABCD into two triangles, for example, triangle ABC and triangle ACD. Step 3: Calculate the area of triangle ABC using the determinant formula: Area = 1/2 |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|. Step 4: For triangle ABC (A(4,2), B(8,2), C(8,14)): Area(ABC) = 1/2 |4(2 - 14) + 8(14 - 2) + 8(2 - 2)| = 1/2 |-48 + 96 + 0| = 1/2 |48| = 24 square units. Step 5: Calculate the area of triangle ACD using the same formula. Step 6: For triangle ACD (A(4,2), C(8,14), D(4,10)): Area(ACD) = 1/2 |4(14 - 10) + 8(10 - 2) + 4(2 - 14)| = 1/2 |16 + 64 - 48| = 1/2 |32| = 16 square units. Step 7: The total area of the quadrilateral ABCD is the sum of the areas of triangle ABC and triangle ACD. Step 8: Area(ABCD) = Area(ABC) + Area(ACD) = 24 + 16 = 40 square units.
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A line passing through points (-2, 5) and (6, b) is perpendicular to the line 20x + 5y = 3. Determine the value of b.
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Solution: Step 1: Find the slope of the given line (L1): 20x + 5y = 3. - Rearrange to slope-intercept form (y = m1x + c1): 5y = -20x + 3 => y = -4x + 3/5. - So, the slope of L1 (m1) = -4. Step 2: Since the line passing through (-2, 5) and (6, b) (L2) is perpendicular to L1, the product of their slopes must be -1. m1 * m2 = -1 => -4 * m2 = -1 => m2 = 1/4. Step 3: Use the two given points for L2, (-2, 5) and (6, b), to calculate its slope (m2) using the slope formula: m = (y2 - y1) / (x2 - x1). - m2 = (b - 5) / (6 - (-2)) = (b - 5) / 8. Step 4: Equate the two expressions for m2 and solve for b: - (b - 5) / 8 = 1/4 - Multiply both sides by 8: b - 5 = 8 * (1/4) => b - 5 = 2. - Solve for b: b = 2 + 5 => b = 7. Step 5: The value of b is 7.
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Determine the equation of a line given its slope is -1/3 and its y-intercept is 6.
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Solution: Step 1: Recall the slope-intercept form of a linear equation: y = mx + c, where m is the slope and c is the y-intercept. Step 2: Identify the given values: Slope (m) = -1/3 Y-intercept (c) = 6 Step 3: Substitute these values into the slope-intercept form: y = (-1/3)x + 6 Step 4: Convert the equation to a standard form (Ax + By = C). Multiply the entire equation by 3 to eliminate the fraction: 3y = -x + 18 Step 5: Rearrange the terms: x + 3y = 18.
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Given that points A(3, -2), B(1, 4), and C(-2, x) are collinear, determine the value of x.
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Solution: Step 1: For three points to be collinear, the slope between any two pairs of points must be equal. Let's calculate the slope of segment AB (m_AB) using points A(3, -2) and B(1, 4). m_AB = (4 - (-2)) / (1 - 3) = (4 + 2) / (-2) = 6 / -2 = -3. Step 2: Calculate the slope of segment BC (m_BC) using points B(1, 4) and C(-2, x). m_BC = (x - 4) / (-2 - 1) = (x - 4) / -3. Step 3: Set m_AB = m_BC since the points are collinear. -3 = (x - 4) / -3. Step 4: Solve for x. (-3) * (-3) = x - 4. 9 = x - 4. x = 9 + 4. x = 13. Alternate Method (using area of triangle = 0): Step 1: For three points (x1, y1), (x2, y2), (x3, y3) to be collinear, the area of the triangle formed by them must be zero. Area = 0.5 * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)| = 0. Step 2: Substitute the coordinates A(3, -2), B(1, 4), C(-2, x). 0 = 3*(4 - x) + 1*(x - (-2)) + (-2)*(-2 - 4). 0 = 3*(4 - x) + 1*(x + 2) + (-2)*(-6). 0 = 12 - 3x + x + 2 + 12. 0 = 26 - 2x. 2x = 26. x = 13.
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Determine the equation of the straight line that passes through the point (3, -5) and has a slope of 2.
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Solution: Step 1: Identify the given point (x1, y1) = (3, -5) and the slope (m) = 2. Step 2: Use the point-slope form of a linear equation: y - y1 = m(x - x1). Step 3: Substitute the given values into the formula: y - (-5) = 2(x - 3) => y + 5 = 2(x - 3). Step 4: Distribute the slope on the right side: y + 5 = 2x - 6. Step 5: Rearrange the equation into the general form (Ax + By + C = 0): 0 = 2x - y - 6 - 5 => 2x - y - 11 = 0. Step 6: The equation of the straight line is 2x - y - 11 = 0.
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Determine the equation of a line that has a slope of -1/2 and passes through the point where the lines x - y = -1 and 3x - 2y = 0 intersect.
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Solution: Step 1: Find the intersection point of the lines x - y = -1 and 3x - 2y = 0. Step 2: From the first equation, x = y - 1. Step 3: Substitute x into the second equation: 3(y - 1) - 2y = 0. Step 4: Solve for y: 3y - 3 - 2y = 0 => y - 3 = 0 => y = 3. Step 5: Substitute y = 3 back into x = y - 1: x = 3 - 1 => x = 2. Step 6: The intersection point is (2, 3). Step 7: The given slope of the required line is m = -1/2. Step 8: Use the point-slope form of a line: y - y1 = m(x - x1), where (x1, y1) = (2, 3) and m = -1/2. Step 9: Substitute the values: y - 3 = (-1/2)(x - 2). Step 10: Multiply both sides by 2: 2(y - 3) = -1(x - 2). Step 11: Simplify: 2y - 6 = -x + 2. Step 12: Rearrange to the standard form: x + 2y = 8.
17
Determine the value of x, given that the line connecting points (-3, 4) and (0, 3) is perpendicular to the line connecting points (5, 7) and (4, x).
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Solution: Step 1: Calculate the slope (m1) of the first line passing through (-3, 4) and (0, 3). m1 = (3 - 4) / (0 - (-3)) = -1 / 3. Step 2: Calculate the slope (m2) of the second line passing through (5, 7) and (4, x). m2 = (x - 7) / (4 - 5) = (x - 7) / -1 = -(x - 7). Step 3: Since the lines are perpendicular, the product of their slopes is -1 (m1 * m2 = -1). (-1/3) * (-(x - 7)) = -1. Step 4: Simplify the equation. (x - 7) / 3 = -1. Step 5: Solve for x. x - 7 = -3. x = -3 + 7. x = 4.
18
Point A (2, 1) divides the line segment BC in the ratio 2 : 3. If the coordinates of B are (1, -3) and C are (4, y), what is the value of y?
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Solution: Step 1: Given point A(2, 1) divides segment BC in the ratio m:n = 2:3. Step 2: Given coordinates of B(x1, y1) = (1, -3). Step 3: Given coordinates of C(x2, y2) = (4, y). Step 4: Use the section formula for the y-coordinate of point A: Ay = (m * y2 + n * y1) / (m + n) Step 5: Substitute the given values: 1 = (2 * y + 3 * (-3)) / (2 + 3) Step 6: Simplify the equation: 1 = (2y - 9) / 5 Step 7: Multiply both sides by 5: 5 * 1 = 2y - 9 5 = 2y - 9 Step 8: Add 9 to both sides: 5 + 9 = 2y 14 = 2y Step 9: Solve for y: y = 14 / 2 = 7. Step 10: The value of y is 7.
19
Determine the coordinates of the points where the graph of the equation 57x - 19y = 399 intersects the x and y coordinate axes.
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Solution: Step 1: To find the x-intercept, set y = 0 in the equation 57x - 19y = 399. 57x - 19*(0) = 399. 57x = 399. x = 399 / 57. x = 7. The x-intercept is at the point (7, 0). Step 2: To find the y-intercept, set x = 0 in the equation 57x - 19y = 399. 57*(0) - 19y = 399. -19y = 399. y = 399 / -19. y = -21. The y-intercept is at the point (0, -21).
20
A line is observed to make an angle (θ) with the positive x-axis such that 90° < θ < 180°. What is the nature of its slope?
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Solution: Step 1: The slope (m) of a line is defined as the tangent of the angle (θ) it makes with the positive x-axis (m = tanθ). Step 2: Given that the angle θ is between 90° and 180° (i.e., in the second quadrant). Step 3: In the second quadrant, the value of tanθ is negative. Step 4: Therefore, the slope of the given line is negative.
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