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The line segment AB has a slope of -2/3. Given the coordinates of point A are (x, -3) and point B are (5, 2), find the value of x.
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Solution: Step 1: Recall the slope formula for a line passing through two points (x1, y1) and (x2, y2): Slope (m) = (y2 - y1) / (x2 - x1). Step 2: Identify the given values: m = -2/3 (x1, y1) = (x, -3) (x2, y2) = (5, 2) Step 3: Substitute the values into the slope formula: -2/3 = (2 - (-3)) / (5 - x) -2/3 = (2 + 3) / (5 - x) -2/3 = 5 / (5 - x) Step 4: Cross-multiply and solve for x: -2 * (5 - x) = 3 * 5 -10 + 2x = 15 2x = 15 + 10 2x = 25 x = 25 / 2 x = 12.5.
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A linear equation is given in the form ax + by + c = 0, where a ≠ 0, b ≠ 0, and c = 0. Through which point does this straight line pass?
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Solution: Step 1: The general form of a linear equation is ax + by + c = 0. Step 2: Given that c = 0, and a ≠ 0, b ≠ 0. Step 3: Substitute c = 0 into the equation: ax + by + 0 = 0, which simplifies to ax + by = 0. Step 4: To find a point that lies on this line, we can test values. Consider the point (0, 0). Step 5: Substitute x = 0 and y = 0 into the equation ax + by = 0: a(0) + b(0) = 0 0 + 0 = 0 0 = 0. Step 6: Since (0, 0) satisfies the equation, the straight line passes through the origin (0, 0).
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Determine the value of k if the line 2x - 3y = 11 is perpendicular to the line 3x + ky = -4.
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Solution: Step 1: Find the slope of the first line (L1): 2x - 3y = 11. - Rearrange to slope-intercept form (y = m1x + c1): -3y = -2x + 11 => y = (2/3)x - 11/3. - So, the slope of L1 (m1) = 2/3. Step 2: Find the slope of the second line (L2): 3x + ky = -4. - Rearrange to slope-intercept form (y = m2x + c2): ky = -3x - 4 => y = (-3/k)x - 4/k. - So, the slope of L2 (m2) = -3/k. Step 3: For two lines to be perpendicular, the product of their slopes must be -1 (m1 * m2 = -1). Step 4: Substitute the slopes into the perpendicularity condition: (2/3) * (-3/k) = -1 => -6 / (3k) = -1 => -2 / k = -1. Step 5: Solve for k: k = -2 / -1 => k = 2. Step 6: The value of k is 2.
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A line passing through (-3, 1) and (x, 5) is parallel to another line passing through (-2, -1) and (6, 3). Determine the value of x.
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Solution: Step 1: Find the slope (m2) of the line passing through points (-2, -1) and (6, 3). m2 = (y2 - y1) / (x2 - x1) = (3 - (-1)) / (6 - (-2)) = (3 + 1) / (6 + 2) = 4 / 8 = 1/2. Step 2: Find the slope (m1) of the line passing through points (-3, 1) and (x, 5). m1 = (y2 - y1) / (x2 - x1) = (5 - 1) / (x - (-3)) = 4 / (x + 3). Step 3: Since the two lines are parallel, their slopes must be equal (m1 = m2). 4 / (x + 3) = 1/2. Step 4: Solve for x: Cross-multiply: 4 * 2 = 1 * (x + 3) 8 = x + 3 x = 8 - 3 x = 5.
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Calculate the area (in square units) of the triangle formed by the graphs of the equations x = 4, y = 3, and 3x + 4y = 12.
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Solution: Step 1: Identify the three lines: L1: x = 4, L2: y = 3, L3: 3x + 4y = 12. Step 2: Find the intersection points of these lines to determine the vertices of the triangle. - Intersection of L1 (x=4) and L2 (y=3): Vertex P1 = (4, 3). - Intersection of L1 (x=4) and L3 (3x + 4y = 12): Substitute x=4 into 3x+4y=12 ⇒ 3(4) + 4y = 12 ⇒ 12 + 4y = 12 ⇒ 4y = 0 ⇒ y = 0. Vertex P2 = (4, 0). - Intersection of L2 (y=3) and L3 (3x + 4y = 12): Substitute y=3 into 3x+4y=12 ⇒ 3x + 4(3) = 12 ⇒ 3x + 12 = 12 ⇒ 3x = 0 ⇒ x = 0. Vertex P3 = (0, 3). Step 3: The vertices of the triangle are (4,3), (4,0), and (0,3). Step 4: Observe that the line x=4 is perpendicular to the line y=3. Therefore, the triangle is a right-angled triangle with the right angle at (4,3). Step 5: The length of the side from (4,0) to (4,3) along the line x=4 is |3 - 0| = 3 units. This can be taken as the base. Step 6: The length of the side from (0,3) to (4,3) along the line y=3 is |4 - 0| = 4 units. This can be taken as the height. Step 7: Calculate the area of the right-angled triangle using the formula (1/2) * base * height = (1/2) * 3 * 4 = 6 square units.
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Calculate the area of the triangle with vertices A(0, 8), O(0, 0), and B(5, 0).
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Solution: Step 1: The given vertices are A(0, 8), O(0, 0), and B(5, 0). Step 2: Observe that vertex O is the origin. Vertex A is on the y-axis (since its x-coordinate is 0), and vertex B is on the x-axis (since its y-coordinate is 0). Step 3: This forms a right-angled triangle with the base along the x-axis and height along the y-axis. Step 4: The length of the base (OB) = |5 - 0| = 5 units. Step 5: The height (OA) = |8 - 0| = 8 units. Step 6: Use the formula for the area of a triangle: Area = (1/2) * base * height. Area = (1/2) * 5 * 8. Area = (1/2) * 40. Area = 20 square units.
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Calculate the area (in square units) of the triangular region bounded by the lines 2x + 5y = 12, x + y = 3, and the x-axis.
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Solution: Step 1: Find the intersection points of the lines with the x-axis (where y = 0). - For 2x + 5y = 12: Set y = 0 => 2x = 12 => x = 6. Point is (6, 0). - For x + y = 3: Set y = 0 => x = 3. Point is (3, 0). Step 2: Find the intersection point of the two lines 2x + 5y = 12 and x + y = 3. - Multiply the second equation by 2: 2(x + y) = 2(3) => 2x + 2y = 6. - Subtract this new equation from the first equation: (2x + 5y = 12) - (2x + 2y = 6) => 3y = 6 => y = 2. - Substitute y = 2 into x + y = 3: x + 2 = 3 => x = 1. - The intersection point is (1, 2). Step 3: The vertices of the triangle are (3, 0), (6, 0), and (1, 2). Step 4: The base of the triangle lies on the x-axis, from x=3 to x=6. Base length = |6 - 3| = 3 units. Step 5: The height of the triangle is the y-coordinate of the third vertex (1, 2), which is 2 units. Step 6: Calculate the area of the triangle using the formula: Area = 1/2 * base * height. Area = 1/2 * 3 * 2 = 3 square units.
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Determine the equation of the line that is perpendicular to the line 2x + 3y = -6 and has a y-intercept of 3.
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Solution: Step 1: Find the slope of the given line 2x + 3y = -6. Rearrange to slope-intercept form (y = mx + c): 3y = -2x - 6 => y = (-2/3)x - 2. So, the slope of the given line (m1) = -2/3. Step 2: Find the slope of the required line, which is perpendicular to the given line. For perpendicular lines, m1 * m2 = -1. (-2/3) * m2 = -1 => m2 = 3/2. Step 3: The required line has a y-intercept of 3. This means c = 3. Step 4: Use the slope-intercept form (y = m2x + c) with m2 = 3/2 and c = 3. y = (3/2)x + 3 Step 5: Convert the equation to a standard form (Ax + By = C). Multiply the entire equation by 2 to eliminate the fraction: 2y = 3x + 6. Step 6: Rearrange the terms: 3x - 2y = -6.
9
The lines defined by the equations 3x + y - 5 = 0 and 2x - y - 5 = 0 intersect at point P(alpha, beta). Calculate the value of (3*alpha + beta).
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Solution: Step 1: Rewrite the equations in standard form: 1) 3x + y = 5 2) 2x - y = 5 Step 2: Solve the system of equations. Add the equations directly: (3x + y) + (2x - y) = 5 + 5 5x = 10 x = 2 Step 3: Substitute x = 2 into either equation (1) or (2) to find y. Using equation (1): 3(2) + y = 5 6 + y = 5 y = 5 - 6 y = -1 Step 4: The intersection point P is (alpha, beta) = (2, -1). So, alpha = 2 and beta = -1. Step 5: Calculate the value of the expression (3*alpha + beta): 3*alpha + beta = 3*(2) + (-1) = 6 - 1 = 5.
10
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).
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A(7, -8) and C(1, 4) are opposite vertices of a square ABCD. Determine the equation of the diagonal BD.
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Solution: Step 1: Understand the properties of a square: its diagonals bisect each other at right angles. This means the diagonals AC and BD are perpendicular, and their midpoint is the same. Step 2: Find the midpoint of diagonal AC, which will also be the midpoint of diagonal BD. Let A=(7, -8) and C=(1, 4). Midpoint (x_m, y_m) = ((x1 + x2)/2, (y1 + y2)/2) x_m = (7 + 1) / 2 = 8 / 2 = 4 y_m = (-8 + 4) / 2 = -4 / 2 = -2 The midpoint of both diagonals is (4, -2). Step 3: Find the slope of diagonal AC. Slope (m_AC) = (y2 - y1) / (x2 - x1) = (4 - (-8)) / (1 - 7) = (4 + 8) / -6 = 12 / -6 = -2. Step 4: Find the slope of diagonal BD. Since diagonals are perpendicular, m_AC * m_BD = -1. -2 * m_BD = -1 => m_BD = 1/2. Step 5: Use the point-slope form (y - y1 = m(x - x1)) for diagonal BD, with the midpoint (4, -2) and slope m_BD = 1/2. y - (-2) = (1/2)(x - 4) y + 2 = (1/2)(x - 4) Step 6: Multiply both sides by 2 to clear fractions: 2(y + 2) = x - 4 2y + 4 = x - 4 Step 7: Rearrange to the standard form: x - 2y = 4 + 4 => x - 2y = 8.
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A line passing through the points (5, a) and (4, 3) is perpendicular to the line x - 6y = 8. Determine the value of 'a'.
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Solution: Step 1: Find the slope of the line x - 6y = 8. Rearrange to slope-intercept form (y = mx + c): -6y = -x + 8 => y = (1/6)x - 8/6. So, the slope of this line (m1) = 1/6. Step 2: Find the slope of the line passing through (5, a) and (4, 3). Let this be m2. m2 = (y2 - y1) / (x2 - x1) = (3 - a) / (4 - 5) = (3 - a) / -1 = a - 3. Step 3: Since the two lines are perpendicular, the product of their slopes is -1 (m1 * m2 = -1). (1/6) * (a - 3) = -1 Step 4: Solve for 'a': a - 3 = -6 a = -6 + 3 a = -3.
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Calculate the area (in square units) enclosed by the lines x = 0, y = 0, x + y = 1, and 2x + 3y = 6.
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Solution: Step 1: Identify the intercepts of the lines on the x and y axes to define two triangles with the origin. For line x + y = 1: - When y = 0, x = 1. Point: (1, 0). - When x = 0, y = 1. Point: (0, 1). This line forms a small triangle with the origin (0,0) and the axes. The vertices are (0,0), (1,0), (0,1). Area of this small triangle (Area_small) = (1/2) * base * height = (1/2) * 1 * 1 = 0.5 square units. For line 2x + 3y = 6: - When y = 0, 2x = 6 ⇒ x = 3. Point: (3, 0). - When x = 0, 3y = 6 ⇒ y = 2. Point: (0, 2). This line forms a large triangle with the origin (0,0) and the axes. The vertices are (0,0), (3,0), (0,2). Area of this large triangle (Area_large) = (1/2) * base * height = (1/2) * 3 * 2 = 3 square units. Step 2: The area bounded by the four given lines (x=0, y=0, x+y=1, 2x+3y=6) is the area of the region between the two lines x+y=1 and 2x+3y=6, in the first quadrant, bounded by the axes. This can be found by subtracting the area of the smaller triangle from the area of the larger triangle. Required Area = Area_large - Area_small. Required Area = 3 - 0.5 = 2.5 square units. In fractional form, this is 5/2 or 2 1/2 square units.
14
Find the coordinates of the centroid for a triangle with vertices A(2, 5), B(-4, 0), and C(5, 4).
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Solution: Step 1: Recall the formula for the centroid of a triangle with vertices (x1, y1), (x2, y2), and (x3, y3): Centroid (X, Y) = ((x1 + x2 + x3)/3, (y1 + y2 + y3)/3) Step 2: Identify the given vertex coordinates: A = (2, 5) => x1 = 2, y1 = 5 B = (-4, 0) => x2 = -4, y2 = 0 C = (5, 4) => x3 = 5, y3 = 4 Step 3: Calculate the x-coordinate of the centroid: X = (2 + (-4) + 5) / 3 = (2 - 4 + 5) / 3 = 3 / 3 = 1 Step 4: Calculate the y-coordinate of the centroid: Y = (5 + 0 + 4) / 3 = 9 / 3 = 3 Step 5: The coordinates of the centroid are (1, 3).
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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.
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Determine the length of the line segment intercepted by the coordinate axes from the graph of the equation 9x - 12y = 108.
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Solution: Step 1: Find the x-intercept by setting y = 0 in the equation 9x - 12y = 108. - 9x - 12(0) = 108 => 9x = 108 => x = 12. The x-intercept point is (12, 0). Step 2: Find the y-intercept by setting x = 0 in the equation 9x - 12y = 108. - 9(0) - 12y = 108 => -12y = 108 => y = -9. The y-intercept point is (0, -9). Step 3: The intercept between the two axes is the line segment connecting the points (12, 0) and (0, -9). Step 4: Use the distance formula to find the length of this segment: Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2). - Distance = sqrt((0 - 12)^2 + (-9 - 0)^2) = sqrt((-12)^2 + (-9)^2). - Distance = sqrt(144 + 81) = sqrt(225). Step 5: Calculate the square root: Distance = 15 units.
17
Which of the given points does the straight line represented by the equation y = 3x always pass through?
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Solution: Step 1: To determine if a point lies on a line, substitute its coordinates (x, y) into the equation of the line. If the equation holds true, the point lies on the line. Step 2: Test the point (0, 0): Substitute x = 0 and y = 0 into the equation y = 3x. 0 = 3(0) 0 = 0. Step 3: Since the equation holds true, the line y = 3x passes through the point (0, 0).
18
Identify which of the following equations represents a line whose graph passes through the origin: 2x - 3y = 3, 2x + 3y = 2, -2x + 3y = 5, or 2x + 3y = 0.
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Solution: Step 1: A line passes through the origin if substituting x=0 and y=0 into its equation results in a true statement (0 = 0). Step 2: Test each given equation with x=0 and y=0: * For 2x - 3y = 3: 2(0) - 3(0) = 0, which is not equal to 3. (Does not pass through origin) * For 2x + 3y = 2: 2(0) + 3(0) = 0, which is not equal to 2. (Does not pass through origin) * For -2x + 3y = 5: -2(0) + 3(0) = 0, which is not equal to 5. (Does not pass through origin) * For 2x + 3y = 0: 2(0) + 3(0) = 0, which is equal to 0. (Passes through origin) Step 3: Therefore, the equation 2x + 3y = 0 represents a line that passes through the origin.
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A line intersects the x-axis at (-3, 0) and the y-axis at (0, 6). Determine the equation of this line.
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Solution: Step 1: The line passes through two points: (x1, y1) = (-3, 0) and (x2, y2) = (0, 6). Step 2: Calculate the slope (m) of the line. m = (y2 - y1) / (x2 - x1) = (6 - 0) / (0 - (-3)) = 6 / 3 = 2. Step 3: Since the y-intercept is given as (0, 6), the value of c in the slope-intercept form (y = mx + c) is 6. Step 4: Substitute the slope m = 2 and y-intercept c = 6 into the slope-intercept form. y = 2x + 6.
20
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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