▶1. Find the distance between points (3, 4) and (7, 7).
Explanation:
To find the distance between two points in the plane, use the distance formula:
Distance = √[(x₂ - x₁)² + (y₂ - y₁)²]
Here, (x₁, y₁) = (3, 4) and (x₂, y₂) = (7, 7).
Calculate the differences: x₂ - x₁ = 7 - 3 = 4, y₂ - y₁ = 7 - 4 = 3.
Now square and add: 4² + 3² = 16 + 9 = 25.
Take the square root: √25 = 5.
Answer: 5 units.
▶2. Find the midpoint of the line segment joining (2, 3) and (4, 5).
Explanation:
The midpoint of a segment is found by averaging the x-coordinates and y-coordinates:
Midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2)
Here, (x₁, y₁) = (2, 3), (x₂, y₂) = (4, 5).
x-coordinate: (2 + 4)/2 = 6/2 = 3
y-coordinate: (3 + 5)/2 = 8/2 = 4
Answer: (3, 4)
▶3. Find the slope of the line passing through (1, 2) and (3, 6).
Explanation:
The slope (m) of a line through two points is:
m = (y₂ - y₁) / (x₂ - x₁)
Here, (x₁, y₁) = (1, 2), (x₂, y₂) = (3, 6).
m = (6 - 2) / (3 - 1) = 4 / 2 = 2
Answer: 2
▶4. If the slope of the line joining (2, y) and (4, 5) is 2, find y.
Explanation:
The slope formula is m = (y₂ - y₁) / (x₂ - x₁).
Here, (x₁, y₁) = (2, y), (x₂, y₂) = (4, 5), and m = 2.
So, 2 = (5 - y) / (4 - 2) = (5 - y) / 2
Multiply both sides by 2: 4 = 5 - y
Rearranged: y = 5 - 4 = 1
Answer: 1
▶5. Find the area of the triangle with vertices (0, 0), (3, 0), and (0, 4).
Explanation:
For a triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃), the area is:
Area = (1/2) × |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|
Plug in the values: (0,0), (3,0), (0,4):
Area = (1/2) × |0(0-4) + 3(4-0) + 0(0-0)|
= (1/2) × |0 + 12 + 0| = (1/2) × 12 = 6
Answer: 6 square units
▶6. Find the point dividing the line segment joining (1, 2) and (6, 7) in the ratio 2:3.
Explanation:
The section formula for dividing a segment in the ratio m:n is:
Point = ((mx₂ + nx₁)/(m+n), (my₂ + ny₁)/(m+n))
Here, m = 2, n = 3, (x₁, y₁) = (1,2), (x₂, y₂) = (6,7):
x = (2×6 + 3×1)/(2+3) = (12+3)/5 = 15/5 = 3
y = (2×7 + 3×2)/5 = (14+6)/5 = 20/5 = 4
Answer: (3, 4)
▶7. Find the equation of the line passing through (2, 3) with slope 4.
Explanation:
The point-slope form of a line is:
y - y₁ = m(x - x₁)
Here, (x₁, y₁) = (2, 3), m = 4:
y - 3 = 4(x - 2)
y - 3 = 4x - 8
y = 4x - 5
Answer: y = 4x - 5
▶8. Find the equation of the line passing through (1, 2) and (3, 4).
Explanation:
First, find the slope: m = (4 - 2)/(3 - 1) = 2/2 = 1
Use point-slope form with (1,2):
y - 2 = 1(x - 1)
y - 2 = x - 1
y = x + 1
Answer: y = x + 1
▶9. Find the equation of the line with slope 3 and y-intercept 5.
Explanation:
The slope-intercept form is y = mx + b.
Here, m = 3, b = 5:
y = 3x + 5
Answer: y = 3x + 5
▶10. Find the equation of the line passing through (0, 3) and (0, 7).
Explanation:
Both points have x = 0, so the line is vertical and passes through x = 0.
Answer: x = 0
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▶11. Find the distance of (3, 4) from the origin.
Explanation:
- Distance from (0, 0): \(\sqrt{(3 - 0)^2 + (4 - 0)^2} = 5\).
Answer: 5 units.
▶12. Find the centroid of the triangle with vertices (1, 2), (3, 4), and (5, 6).
Explanation:
- Centroid formula: \(\left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)\).
- Plug in: \(x_{\text{cent}} = 3\), \(y_{\text{cent}} = 4\).
Answer: (3, 4).
▶13. Are points (1, 1), (2, 2), and (3, 3) collinear?
Explanation:
- Area formula: \(\frac{1}{2} \left| 1(2 - 3) + 2(3 - 1) + 3(1 - 2) \right| = 0\).
- Since area is 0, points are collinear.
Answer: Yes.
▶14. Find the equation of the circle with center (2, 3) and radius 4.
Explanation:
- Standard form: \((x - 2)^2 + (y - 3)^2 = 16\).
Answer: \((x - 2)^2 + (y - 3)^2 = 16\).
▶15. Find the center and radius of \(x^2 + y^2 - 4x - 6y + 9 = 0\).
Explanation:
- Complete the square: \((x - 2)^2 + (y - 3)^2 = 4\).
- Center: (2, 3), radius: 2.
Answer: Center (2, 3), radius 2.
▶16. Find the distance between \(3x + 4y = 10\) and \(3x + 4y = 20\).
Explanation:
- Distance formula: \(\frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}} = 2\).
Answer: 2 units.
▶17. Find the intersection of \(2x + 3y = 7\) and \(4x + 5y = 13\).
Explanation:
- Solve the system: Multiply first by 2: \(4x + 6y = 14\). Subtract second: \(y = 1\). Substitute: \(x = 2\).
Answer: (2, 1).
▶18. Find the angle between lines \(y = 2x + 3\) and \(y = 3x + 5\).
Explanation:
- Slopes: \(m_1 = 2\), \(m_2 = 3\).
- \(\tan \theta = |(3-2)/(1+2\cdot3)| = 1/7\).
Answer: \(\tan^{-1}(1/7)\).
▶19. Find the reflection of (2, 3) in the x-axis.
Explanation:
- Reflection in x-axis: y-coordinate changes sign. (2, 3) → (2, -3).
Answer: (2, -3).
▶20. Find the area of the quadrilateral with vertices (0, 0), (1, 1), (2, 0), and (1, -1).
Explanation:
- Use shoelace formula. Area = 2.
Answer: 2 square units.
▶21. Find the tangent to \(x^2 + y^2 = 25\) at (3, 4).
Explanation:
- Tangent formula: \(xx_1 + yy_1 = r^2\). Plug in: \(x_1 = 3\), \(y_1 = 4\), \(r^2 = 25\).
Answer: \(3x + 4y = 25\).
▶22. Find the length of the tangent from (4, 5) to \(x^2 + y^2 - 2x - 4y - 4 = 0\).
Explanation:
- Tangent length: \(\sqrt{S_1}\), where \(S_1 = x_1^2 + y_1^2 - 2x_1 - 4y_1 - 4\). Plug in: \(S_1 = 9\). Length = 3.
Answer: 3 units.
▶23. Find the circle passing through (1, 0), (0, 1), and (0, 0).
Explanation:
- Let circle equation: \(x^2 + y^2 + Dx + Ey + F = 0\). Plug in points to solve: \(D = -1\), \(E = -1\), \(F = 0\).
Answer: \(x^2 + y^2 - x - y = 0\).
▶24. Find the distance from (2, 3) to \(3x + 4y + 5 = 0\).
Explanation:
- Distance formula: \(\frac{|3(2) + 4(3) + 5|}{5} = 4.6\).
Answer: 4.6 units.
▶25. Find the chord length cut by \(3x + 4y = 15\) on \(x^2 + y^2 = 25\).
Explanation:
- Distance from center (0,0) to line: 3. Radius \(r = 5\). Half chord length = 4. Full length = 8.
Answer: 8 units.
▶26. Find the line through (1, 1) perpendicular to \(3x + 4y = 5\).
Explanation:
- Slope of given line: \(-3/4\). Perpendicular slope: \(4/3\). Equation: \(y - 1 = \frac{4}{3}(x - 1)\) → \(4x - 3y = 1\).
Answer: \(4x - 3y = 1\).
▶27. Find the circle with diameter endpoints (1, 2) and (3, 4).
Explanation:
- Center: midpoint = (2, 3). Radius: \(\sqrt{2}\). Equation: \((x - 2)^2 + (y - 3)^2 = 2\).
Answer: \((x - 2)^2 + (y - 3)^2 = 2\).
▶28. Find the area of the triangle formed by \(x = 0\), \(y = 0\), and \(3x + 4y = 12\).
Explanation:
- Intercepts: (4, 0), (0, 3). Area = 6.
Answer: 6 square units.
▶29. Find the equation of the circle touching both axes with radius 5.
Explanation:
- Center: (±5, ±5). Possible circles: \((x - 5)^2 + (y - 5)^2 = 25\), etc.
Answer: Four possible circles (e.g., \((x - 5)^2 + (y - 5)^2 = 25\)).
▶30. Find the foot of the perpendicular from (1, 2) to 3x + 4y + 5 = 0.
Explanation:
- The line perpendicular to 3x + 4y + 5 = 0 has slope 4/3. The equation from (1,2) is y - 2 = (4/3)(x - 1). Solve this with the original line to get the foot of the perpendicular: x = -23/25, y = -14/25.
Answer: (-23/25, -14/25).
▶31. Find the distance in 3D between (1, 2, 3) and (4, 5, 6).
Explanation:
- The distance between two points in 3D, (x₁, y₁, z₁) and (x₂, y₂, z₂), is given by the formula:
\(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\)
- Substitute the values: (1,2,3) and (4,5,6):
\(\sqrt{(4-1)^2 + (5-2)^2 + (6-3)^2} = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{9 + 9 + 9} = \sqrt{27} = 3\sqrt{3}\)
Answer: 3√3 units.
▶32. Find the midpoint of (1, 2, 3) and (3, 4, 5) in 3D.
Explanation:
- The midpoint formula in 3D is:
\(\left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right)\)
- Substitute the values: (1,2,3) and (3,4,5):
\(\left( \frac{1+3}{2}, \frac{2+4}{2}, \frac{3+5}{2} \right) = (2, 3, 4)\)
Answer: (2, 3, 4).
▶33. Find the equation of the sphere with center (1, 2, 3) and radius 4.
Explanation:
- The equation of a sphere with center (h, k, l) and radius r is:
\((x - h)^2 + (y - k)^2 + (z - l)^2 = r^2\)
- Substitute h=1, k=2, l=3, r=4:
\((x - 1)^2 + (y - 2)^2 + (z - 3)^2 = 16\)
Answer: (x - 1)² + (y - 2)² + (z - 3)² = 16.
▶34. Find the distance from (1, 2, 3) to the origin in 3D.
Explanation:
- Use the 3D distance formula from (x, y, z) to (0, 0, 0):
\(\sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}\)
Answer: √14 units.
▶35. Find the point dividing (1, 2, 3) and (4, 5, 6) in the ratio 2:1 in 3D.
Explanation:
- The section formula in 3D for dividing (x₁, y₁, z₁) and (x₂, y₂, z₂) in the ratio m:n is:
\(\left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n} \right)\)
- Here, m=2, n=1, (x₁, y₁, z₁) = (1,2,3), (x₂, y₂, z₂) = (4,5,6):
x = (2×4 + 1×1)/3 = 3, y = (2×5 + 1×2)/3 = 4, z = (2×6 + 1×3)/3 = 5
Answer: (3, 4, 5).
▶36. Find the angle between lines with direction ratios (1, 2, 3) and (4, 5, 6) in 3D.
Explanation:
- The angle θ between two lines with direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂) is given by:
\(\cos \theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}}\)
- Substitute: (1,2,3) and (4,5,6):
Numerator: 1×4 + 2×5 + 3×6 = 4 + 10 + 18 = 32
Denominator: sqrt(1+4+9) × sqrt(16+25+36) = sqrt(14) × sqrt(77)
- So, \(\cos \theta = \frac{32}{\sqrt{14}\sqrt{77}}\)
Answer: θ = cos⁻¹(32 / (√14 × √77))
▶37. Find the area of the triangle with vertices (1, 0, 0), (0, 1, 0), and (0, 0, 1) in 3D.
Explanation:
- First, form two vectors from the points: A = (1,0,0), B = (0,1,0), C = (0,0,1)
- Vector AB = B - A = (-1, 1, 0), Vector AC = C - A = (-1, 0, 1)
- The area of the triangle is half the magnitude of the cross product of AB and AC.
- AB × AC = (1, 1, 1), |AB × AC| = √(1² + 1² + 1²) = √3
- Area = (1/2) × √3 = √3/2
Answer: √3/2 square units.
▶38. Find the plane through (1, 2, 3) with normal vector (4, 5, 6).
Explanation:
- The equation of a plane with normal vector (a, b, c) passing through (x₁, y₁, z₁) is:
a(x - x₁) + b(y - y₁) + c(z - z₁) = 0
- Substitute a=4, b=5, c=6, x₁=1, y₁=2, z₁=3:
4(x-1) + 5(y-2) + 6(z-3) = 0
- Expand: 4x - 4 + 5y - 10 + 6z - 18 = 0 → 4x + 5y + 6z - 32 = 0
Answer: 4x + 5y + 6z = 32
▶39. Find the distance from (1, 2, 3) to 2x + 3y + 4z + 5 = 0 in 3D.
Explanation:
- The distance from point (x₁, y₁, z₁) to plane ax + by + cz + d = 0 is:
\(\frac{|a x_1 + b y_1 + c z_1 + d|}{\sqrt{a^2 + b^2 + c^2}}\)
- Substitute a=2, b=3, c=4, d=5, (x₁, y₁, z₁) = (1,2,3):
Numerator: |2×1 + 3×2 + 4×3 + 5| = |2 + 6 + 12 + 5| = |25| = 25
Denominator: sqrt(2² + 3² + 4²) = sqrt(4 + 9 + 16) = sqrt(29)
- Distance = 25 / √29
Answer: 25/√29 units.
▶40. Find the equation of the line through (1, 2, 3) and (4, 5, 6) in 3D.
Explanation:
- The direction ratios are (4-1, 5-2, 6-3) = (3, 3, 3) or (1, 1, 1).
- The symmetric form of the line is:
\(\frac{x - 1}{1} = \frac{y - 2}{1} = \frac{z - 3}{1}\)
Answer: (x - 1)/1 = (y - 2)/1 = (z - 3)/1
▶41. Where does (x - 1)/2 = (y - 2)/3 = (z - 3)/4 meet x + y + z = 10?
Explanation:
- Let the parameter be k. Then x = 2k + 1, y = 3k + 2, z = 4k + 3.
- Substitute into the plane equation: (2k+1) + (3k+2) + (4k+3) = 10
- 2k + 3k + 4k + 1 + 2 + 3 = 10 → 9k + 6 = 10 → 9k = 4 → k = 4/9
- Substitute k back:
x = 2×4/9 + 1 = 8/9 + 1 = 17/9
y = 3×4/9 + 2 = 12/9 + 2 = 10/3
z = 4×4/9 + 3 = 16/9 + 3 = 43/9
Answer: (17/9, 10/3, 43/9)
▶42. Find the angle between planes 2x + 3y + 4z = 5 and 3x + 4y + 5z = 6.
Explanation:
- The angle between two planes is the angle between their normal vectors.
- The normal vector of the first plane is (2, 3, 4), and for the second plane it is (3, 4, 5).
- The formula for the angle θ between two vectors a and b is:
\(\cos \theta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}\)
- Calculate the dot product: 2×3 + 3×4 + 4×5 = 6 + 12 + 20 = 38
- Find the magnitudes:
|a| = sqrt(2² + 3² + 4²) = sqrt(4 + 9 + 16) = sqrt(29)
|b| = sqrt(3² + 4² + 5²) = sqrt(9 + 16 + 25) = sqrt(50)
- Plug into the formula:
cos θ = 38 / (sqrt(29) × sqrt(50)) = 38 / (5√58)
Answer: θ = cos⁻¹(38 / (5√58))
▶43. Find the sphere with diameter endpoints (1, 2, 3) and (5, 6, 7).
Explanation:
- Center: (3, 4, 5). Radius: \(2\sqrt{3}\). Equation: \((x - 3)^2 + (y - 4)^2 + (z - 5)^2 = 12\).
Answer: \((x - 3)^2 + (y - 4)^2 + (z - 5)^2 = 12\).
▶44. Find the tangent plane to \(x^2 + y^2 + z^2 = 25\) at (3, 4, 0).
Explanation:
- Tangent plane: \(3x + 4y = 25\).
Answer: \(3x + 4y = 25\).
▶45. Find the shortest distance between \(\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}\) and \(\frac{x-2}{3} = \frac{y-3}{4} = \frac{z-4}{5}\).
Explanation:
- Lines: L1 = (1,2,3) + s(2,3,4), L2 = (2,3,4) + t(3,4,5).
- Vector between points: (1,1,1).
- Cross product: (-1,2,-1).
- Shortest distance = 0 (lines intersect).
Answer: 0 (lines intersect).
▶46. Find the area of the parallelogram with vectors (1, 2, 3) and (4, 5, 6).
Explanation:
- Cross product: (-3, 6, -3). Magnitude: \(3\sqrt{6}\).
Answer: \(3\sqrt{6}\) square units.
▶47. Find the distance between (0, 0) and (a, b) in 2D.
Explanation:
- \(\sqrt{a^2 + b^2}\).
Answer: \(\sqrt{a^2 + b^2}\).
▶48. What is the slope of the line x = 5?
Explanation:
- Vertical lines have undefined slope.
Answer: Undefined.
▶49. What is the equation of the x-axis?
Explanation:
- All points on the x-axis have y = 0.
Answer: y = 0.
▶50. Are y = 2x + 3 and y = 2x - 4 parallel?
Explanation:
- Both have slope 2, so yes.
Answer: Yes.
▶51. Are y = 3x + 2 and y = -1/3x + 5 perpendicular?
Explanation:
- Product of slopes: 3 × (-1/3) = -1, so yes.
Answer: Yes.
▶52. Find the area of the circle x^2 + y^2 = 36.
Explanation:
- Radius r = 6, area = 36π.
Answer: 36π square units.
▶53. Find the circumference of (x-1)^2 + (y-2)^2 = 9.
Explanation:
- Radius r = 3, circumference = 6π.
Answer: 6π units.
▶54. Is (3, 4) on x^2 + y^2 = 25?
Explanation:
- Plug in: 9 + 16 = 25, so yes.
Answer: Yes.
▶55. How many points do x^2 + y^2 = 25 and x + y = 5 intersect at?
Explanation:
- Substitute y = 5 - x into circle: x^2 + (5 - x)^2 = 25 → 2x^2 - 10x = 0 → x = 0 or x = 5.
- Points: (0,5), (5,0).
Answer: 2 points.
▶56. What is the center of x^2 + y^2 - 6x - 8y = 0?
Explanation:
- Complete the square: (x-3)^2 + (y-4)^2 = 25.
Answer: (3, 4).
▶57. What is the radius of x^2 + y^2 + 4x - 6y - 3 = 0?
Explanation:
- Complete the square: (x+2)^2 + (y-3)^2 = 16.
Answer: 4.
▶58. Find the line with x-intercept 3 and y-intercept 4.
Explanation:
- Intercept form: x/3 + y/4 = 1. Multiply by 12: 4x + 3y = 12.
Answer: 4x + 3y = 12.
▶59. Find the perpendicular bisector of (1, 2) and (3, 4).
Explanation:
- Midpoint: (2, 3). Slope: 1. Perpendicular slope: -1. Equation: y - 3 = -1(x - 2) → y = -x + 5.
Answer: y = -x + 5.
▶60. Find the area of the rectangle with vertices (0,0), (5,0), (5,3), (0,3).
Explanation:
- Length = 5, width = 3, area = 15.
Answer: 15 square units.
▶61. Find the distance between centers of x^2 + y^2 = 4 and x^2 + y^2 - 4x - 4y + 4 = 0.
Explanation:
- First center: (0,0).
- Second circle: (x-2)^2 + (y-2)^2 = 4, center (2,2).
- Distance: sqrt((2-0)^2 + (2-0)^2) = sqrt(8) = 2sqrt(2).
Answer: 2√2 units.
▶62. Find the reflection of (3,4) in the y-axis.
Explanation:
- Reflection in y-axis: x-coordinate changes sign.
- So, (3,4) → (-3,4).
Answer: (-3, 4).
▶63. Find the reflection of (1,2) in the origin.
Explanation:
- Both coordinates change sign: (1,2) → (-1,-2).
Answer: (-1, -2).
▶64. Find the reflection of (4,5) in y=x.
Explanation:
- Swap x and y: (4,5) → (5,4).
Answer: (5, 4).
▶65. Find the point symmetric to (3,7) with respect to (1,1).
Explanation:
- (1,1) is midpoint of (3,7) and (x,y).
- (3+x)/2 = 1 → x = -1, (7+y)/2 = 1 → y = -5.
Answer: (-1, -5).
▶66. Find the horizontal line through (3,4).
Explanation:
- Horizontal lines have constant y, so y = 4.
Answer: y = 4.
▶67. Find the vertical line through (5,6).
Explanation:
- Vertical lines have constant x, so x = 5.
Answer: x = 5.
▶68. Find the circumference of x^2 + y^2 - 6x - 8y = 0.
Explanation:
- From earlier, radius = 5, circumference = 2π × 5 = 10π.
Answer: 10π units.
▶69. Find the line with slope -2 and y-intercept 3.
Explanation:
- y = mx + b → y = -2x + 3.
Answer: y = -2x + 3.
▶70. Find the centroid of the tetrahedron with vertices (1,2,3), (4,5,6), (7,8,9), (10,11,12).
Explanation:
- Centroid: ((1+4+7+10)/4, (2+5+8+11)/4, (3+6+9+12)/4) = (5.5, 6.5, 7.5).
Answer: (5.5, 6.5, 7.5).