Discover the answers you need at Westonci.ca, where experts provide clear and concise information on various topics. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
### (i) Point Equidistant from (2, 3) and (4, 5)
To find the point which is always equidistant from (2, 3) and (4, 5), we need to determine the perpendicular bisector of the line segment joining these points.
1. Coordinates of the Points:
- Point [tex]\( A = (2, 3) \)[/tex]
- Point [tex]\( B = (4, 5) \)[/tex]
2. Midpoint of Segment AB:
The midpoint [tex]\( M \)[/tex] of the segment joining points [tex]\( A \)[/tex] and [tex]\( B \)[/tex] is calculated as:
[tex]\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{2 + 4}{2}, \frac{3 + 5}{2} \right) = (3, 4) \][/tex]
3. Slope of Segment AB:
The slope [tex]\( m_{AB} \)[/tex] of the line segment joining [tex]\( A \)[/tex] and [tex]\( B \)[/tex] is calculated as:
[tex]\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 3}{4 - 2} = 1 \][/tex]
4. Slope of the Perpendicular Bisector:
The slope of the perpendicular bisector of the segment is the negative reciprocal of [tex]\( m_{AB} \)[/tex]:
[tex]\[ m_{PB} = -\frac{1}{m_{AB}} = -1 \][/tex]
5. Equation of the Perpendicular Bisector:
The equation of the line in point-slope form is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
For the perpendicular bisector passing through midpoint [tex]\( M(3, 4) \)[/tex]:
[tex]\[ y - 4 = -1(x - 3) \][/tex]
Simplifying this, we get:
[tex]\[ y = -x + 7 \][/tex]
### (ii) Distance from Y-axis same as Distance from Point (1, 0)
Let [tex]\( P(x, y) \)[/tex] be a point whose distance from the Y-axis is the same as its distance from the point (1, 0).
1. Distance from Y-axis:
The distance of point [tex]\( P(x, y) \)[/tex] from the Y-axis is [tex]\( |x| \)[/tex].
2. Distance from Point (1, 0):
The distance of point [tex]\( P(x, y) \)[/tex] from the point (1, 0) is:
[tex]\[ \sqrt{(x-1)^2 + y^2} \][/tex]
3. Equating Distances:
According to the problem, these distances are equal:
[tex]\[ |x| = \sqrt{(x-1)^2 + y^2} \][/tex]
Squaring both sides, we get:
[tex]\[ x^2 = (x-1)^2 + y^2 \][/tex]
Expanding and simplifying:
[tex]\[ x^2 = x^2 - 2x + 1 + y^2 \][/tex]
[tex]\[ 0 = -2x + 1 + y^2 \][/tex]
[tex]\[ 2x = 1 + y^2 \][/tex]
[tex]\[ x = \frac{1 + y^2}{2} \][/tex]
### (iii) Point P Moves such that its Distance from (1, 0) is Twice its Distance from (0, 0)
Let [tex]\( P(x, y) \)[/tex] be a point whose distance from the point (1, 0) is twice the distance from the point (0, 0).
1. Distance from (1, 0):
[tex]\[ \sqrt{(x-1)^2 + y^2} \][/tex]
2. Distance from (0, 0):
[tex]\[ \sqrt{x^2 + y^2} \][/tex]
3. Relation between Distances:
[tex]\[ \sqrt{(x-1)^2 + y^2} = 2 \cdot \sqrt{x^2 + y^2} \][/tex]
Squaring both sides, we get:
[tex]\[ (x-1)^2 + y^2 = 4(x^2 + y^2) \][/tex]
Expanding and simplifying:
[tex]\[ x^2 - 2x + 1 + y^2 = 4x^2 + 4y^2 \][/tex]
[tex]\[ 1 - 2x = 3x^2 + 3y^2 \][/tex]
Moving all terms to one side yields:
[tex]\[ 3x^2 + 3y^2 + 2x - 1 = 0 \][/tex]
In summary, the results are:
- The midpoint coordinates of the line segment joining (2, 3) and (4, 5) are [tex]\((3.0, 4.0)\)[/tex].
- The equation of the perpendicular bisector is [tex]\( y = -x + 7 \)[/tex].
- The equation of the locus of a point whose distance from the Y-axis is the same as its distance from (1, 0) is [tex]\( x = \frac{1 + y^2}{2} \)[/tex].
- The equation of the locus of a point whose distance from (1, 0) is twice its distance from (0, 0) is [tex]\( 3x^2 + 3y^2 + 2x - 1 = 0 \)[/tex].
To find the point which is always equidistant from (2, 3) and (4, 5), we need to determine the perpendicular bisector of the line segment joining these points.
1. Coordinates of the Points:
- Point [tex]\( A = (2, 3) \)[/tex]
- Point [tex]\( B = (4, 5) \)[/tex]
2. Midpoint of Segment AB:
The midpoint [tex]\( M \)[/tex] of the segment joining points [tex]\( A \)[/tex] and [tex]\( B \)[/tex] is calculated as:
[tex]\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{2 + 4}{2}, \frac{3 + 5}{2} \right) = (3, 4) \][/tex]
3. Slope of Segment AB:
The slope [tex]\( m_{AB} \)[/tex] of the line segment joining [tex]\( A \)[/tex] and [tex]\( B \)[/tex] is calculated as:
[tex]\[ m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{5 - 3}{4 - 2} = 1 \][/tex]
4. Slope of the Perpendicular Bisector:
The slope of the perpendicular bisector of the segment is the negative reciprocal of [tex]\( m_{AB} \)[/tex]:
[tex]\[ m_{PB} = -\frac{1}{m_{AB}} = -1 \][/tex]
5. Equation of the Perpendicular Bisector:
The equation of the line in point-slope form is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
For the perpendicular bisector passing through midpoint [tex]\( M(3, 4) \)[/tex]:
[tex]\[ y - 4 = -1(x - 3) \][/tex]
Simplifying this, we get:
[tex]\[ y = -x + 7 \][/tex]
### (ii) Distance from Y-axis same as Distance from Point (1, 0)
Let [tex]\( P(x, y) \)[/tex] be a point whose distance from the Y-axis is the same as its distance from the point (1, 0).
1. Distance from Y-axis:
The distance of point [tex]\( P(x, y) \)[/tex] from the Y-axis is [tex]\( |x| \)[/tex].
2. Distance from Point (1, 0):
The distance of point [tex]\( P(x, y) \)[/tex] from the point (1, 0) is:
[tex]\[ \sqrt{(x-1)^2 + y^2} \][/tex]
3. Equating Distances:
According to the problem, these distances are equal:
[tex]\[ |x| = \sqrt{(x-1)^2 + y^2} \][/tex]
Squaring both sides, we get:
[tex]\[ x^2 = (x-1)^2 + y^2 \][/tex]
Expanding and simplifying:
[tex]\[ x^2 = x^2 - 2x + 1 + y^2 \][/tex]
[tex]\[ 0 = -2x + 1 + y^2 \][/tex]
[tex]\[ 2x = 1 + y^2 \][/tex]
[tex]\[ x = \frac{1 + y^2}{2} \][/tex]
### (iii) Point P Moves such that its Distance from (1, 0) is Twice its Distance from (0, 0)
Let [tex]\( P(x, y) \)[/tex] be a point whose distance from the point (1, 0) is twice the distance from the point (0, 0).
1. Distance from (1, 0):
[tex]\[ \sqrt{(x-1)^2 + y^2} \][/tex]
2. Distance from (0, 0):
[tex]\[ \sqrt{x^2 + y^2} \][/tex]
3. Relation between Distances:
[tex]\[ \sqrt{(x-1)^2 + y^2} = 2 \cdot \sqrt{x^2 + y^2} \][/tex]
Squaring both sides, we get:
[tex]\[ (x-1)^2 + y^2 = 4(x^2 + y^2) \][/tex]
Expanding and simplifying:
[tex]\[ x^2 - 2x + 1 + y^2 = 4x^2 + 4y^2 \][/tex]
[tex]\[ 1 - 2x = 3x^2 + 3y^2 \][/tex]
Moving all terms to one side yields:
[tex]\[ 3x^2 + 3y^2 + 2x - 1 = 0 \][/tex]
In summary, the results are:
- The midpoint coordinates of the line segment joining (2, 3) and (4, 5) are [tex]\((3.0, 4.0)\)[/tex].
- The equation of the perpendicular bisector is [tex]\( y = -x + 7 \)[/tex].
- The equation of the locus of a point whose distance from the Y-axis is the same as its distance from (1, 0) is [tex]\( x = \frac{1 + y^2}{2} \)[/tex].
- The equation of the locus of a point whose distance from (1, 0) is twice its distance from (0, 0) is [tex]\( 3x^2 + 3y^2 + 2x - 1 = 0 \)[/tex].
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.
