Beshale General Secondary School Second Semester Mathematics Worksheet 11 (2017 F.C.)Work Out The Following Problems Accordingly:1. Let \[$ A = (2,3) \$\] And \[$ B = (3,5) \$\]. Then Find: A) The Position Vector Of \[$ AB
Beshale General Secondary School Second Semester Mathematics Worksheet 11 (2017 F.C.)
Solution to Problems
Problem 1: Position Vector of AB
Let { A = (2,3) $}$ and { B = (3,5) $}$. Then find:
a) The position vector of { AB
To find the position vector of [$ AB, we need to subtract the coordinates of point A from the coordinates of point B.
The position vector of [$ AB is given by:
[$ \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} $}$
where { \overrightarrow{OA} $}$ is the position vector of point A and { \overrightarrow{OB} $}$ is the position vector of point B.
Since { A = (2,3) $}$ and { B = (3,5) $}$, we have:
{ \overrightarrow{OA} = 2\hat{i} + 3\hat{j} $}$
and
{ \overrightarrow{OB} = 3\hat{i} + 5\hat{j} $}$
Therefore,
{ \overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} $}$
{ = (3\hat{i} + 5\hat{j}) - (2\hat{i} + 3\hat{j}) $}$
{ = \hat{i} + 2\hat{j} $}$
Hence, the position vector of { AB is \hat{i} + 2\hat{j} $}$.
Problem 2: Midpoint of AB
Find the midpoint of { AB.
The midpoint of [$ AB is given by:
[$ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) $}$
where { (x_1, y_1) $}$ are the coordinates of point A and { (x_2, y_2) $}$ are the coordinates of point B.
Since { A = (2,3) $}$ and { B = (3,5) $}$, we have:
{ M = \left( \frac{2 + 3}{2}, \frac{3 + 5}{2} \right) $}$
{ = \left( \frac{5}{2}, 4 \right) $}$
Hence, the midpoint of { AB is \left( \frac{5}{2}, 4 \right) $}$.
Problem 3: Distance Between A and B
Find the distance between points A and B.
The distance between points A and B is given by:
{ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $}$
where { (x_1, y_1) $}$ are the coordinates of point A and { (x_2, y_2) $}$ are the coordinates of point B.
Since { A = (2,3) $}$ and { B = (3,5) $}$, we have:
{ AB = \sqrt{(3 - 2)^2 + (5 - 3)^2} $}$
{ = \sqrt{1^2 + 2^2} $}$
{ = \sqrt{5} $}$
Hence, the distance between points A and B is \sqrt{5} $].
Problem 4: Slope of AB
Find the slope of { AB.
The slope of [$ AB is given by:
[$ m = \frac{y_2 - y_1}{x_2 - x_1} $}$
where { (x_1, y_1) $}$ are the coordinates of point A and { (x_2, y_2) $}$ are the coordinates of point B.
Since { A = (2,3) $}$ and { B = (3,5) $}$, we have:
{ m = \frac{5 - 3}{3 - 2} $}$
{ = \frac{2}{1} $}$
{ = 2 $}$
Hence, the slope of { AB is 2 $}$.
Problem 5: Equation of AB
Find the equation of { AB.
The equation of [$ AB is given by:
[$ y - y_1 = m(x - x_1) $}$
where { (x_1, y_1) $}$ are the coordinates of point A and { m $}$ is the slope of { AB.
Since [$ A = (2,3) $}$ and the slope of { AB is 2 $}$, we have:
{ y - 3 = 2(x - 2) $}$
{ y - 3 = 2x - 4 $}$
{ y = 2x - 1 $}$
Hence, the equation of { AB is y = 2x - 1 $}$.
Problem 6: Perpendicular Line
Find the equation of a line perpendicular to { AB and passing through point A.
The equation of a line perpendicular to [$ AB is given by:
[$ y - y_1 = -\frac{1}{m}(x - x_1) $}$
where { (x_1, y_1) $}$ are the coordinates of point A and { m $}$ is the slope of { AB.
Since [$ A = (2,3) $}$ and the slope of { AB is 2 $}$, we have:
{ y - 3 = -\frac{1}{2}(x - 2) $}$
{ y - 3 = -\frac{1}{2}x + 1 $}$
{ y = -\frac{1}{2}x + 4 $}$
Hence, the equation of a line perpendicular to { AB and passing through point A is y = -\frac{1}{2}x + 4 $}$.
Problem 7: Angle Between AB and x-axis
Find the angle between { AB and the x-axis.
The angle between [$ AB and the x-axis is given by:
[$ \theta = \tan^{-1} \left( \frac{m}{1} \right) $}$
where { m $}$ is the slope of { AB.
Since the slope of [$ AB is 2 $}$, we have:
{ \theta = \tan^{-1} (2) $}$
{ \theta = 63.43^{\circ} $}$
Hence, the angle between { AB and the x-axis is 63.43^{\circ} $}$.
Problem 8: Angle Between AB and y-axis
Find the angle between { AB and the y-axis.
The angle between [$ AB and the y-axis is given by:
[$ \theta = \tan^{-1} \left( \frac{1}{m} \right) $}$
where { m $}$ is the slope of { AB.
Since the slope of [$ AB is 2 $}$, we have:
{ \theta = \tan^{-1} \left( \frac{1}{2} \right) $}$
{ \theta = 26.57^{\circ} $}$
Hence, the angle between { AB and the y-axis is 26.57^{\circ} $}$.
Problem 9: Length of AB
Find the length of { AB.
The length of [$ AB is given by:
[$ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $}$
where { (x_1, y_1) $}$ are the coordinates of point A and { (x_2, y_2) $}$ are the coordinates of point B.
Since { A = (2,3) $}$ and { B = (3,5) $}$, we have:
{ AB = \sqrt{(3 - 2)^2 + (5 - 3)^2} $}$
{ = \sqrt{1^2 + 2^2} $}$
{ = \sqrt{5} $}$
Hence, the length of { AB is \sqrt{5} $}$.
Problem 10: Area of Triangle OAB
Find the area of triangle OAB.
The area of triangle OAB is given by:
{ \frac{1}{2} \times \text{base} \times \text{height} $}$
where the base is the length of { AB and the height is the distance from point O to the line [$ AB.
Since the length of [$ AB is \sqrt{5} $}$ and the distance from point O to the line { AB is 2 $}$, we have:
{ \text{Area} = \frac{1}{2} \times \sqrt{5} \times 2 $}$
[$ = \sqrt{5} {{content}}lt;br/> Beshale General Secondary School Second Semester Mathematics Worksheet 11 (2017 F.C.)
Q&A Section
Q1: What is the position vector of AB?
A1: The position vector of AB is given by \hat{i} + 2\hat{j}.
Q2: How do you find the midpoint of AB?
A2: The midpoint of AB is given by \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right), where (x_1, y_1) are the coordinates of point A and (x_2, y_2) are the coordinates of point B.
Q3: What is the distance between points A and B?
A3: The distance between points A and B is given by \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}, where (x_1, y_1) are the coordinates of point A and (x_2, y_2) are the coordinates of point B.
Q4: What is the slope of AB?
A4: The slope of AB is given by \frac{y_2 - y_1}{x_2 - x_1}, where (x_1, y_1) are the coordinates of point A and (x_2, y_2) are the coordinates of point B.
Q5: What is the equation of AB?
A5: The equation of AB is given by y = 2x - 1.
Q6: How do you find the equation of a line perpendicular to AB and passing through point A?
A6: The equation of a line perpendicular to AB is given by y - y_1 = -\frac{1}{m}(x - x_1), where (x_1, y_1) are the coordinates of point A and m is the slope of AB.
Q7: What is the angle between AB and the x-axis?
A7: The angle between AB and the x-axis is given by \theta = \tan^{-1} \left( \frac{m}{1} \right), where m is the slope of AB.
Q8: What is the angle between AB and the y-axis?
A8: The angle between AB and the y-axis is given by \theta = \tan^{-1} \left( \frac{1}{m} \right), where m is the slope of AB.
Q9: What is the length of AB?
A9: The length of AB is given by \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}, where (x_1, y_1) are the coordinates of point A and (x_2, y_2) are the coordinates of point B.
Q10: What is the area of triangle OAB?
A10: The area of triangle OAB is given by \frac{1}{2} \times \text{base} \times \text{height}, where the base is the length of AB and the height is the distance from point O to the line AB.
Q11: How do you find the coordinates of the point where the line AB intersects the x-axis?
A11: To find the coordinates of the point where the line AB intersects the x-axis, we need to set y = 0 in the equation of the line AB and solve for x.
Q12: How do you find the coordinates of the point where the line AB intersects the y-axis?
A12: To find the coordinates of the point where the line AB intersects the y-axis, we need to set x = 0 in the equation of the line AB and solve for y.
Q13: What is the slope of the line perpendicular to AB?
A13: The slope of the line perpendicular to AB is given by -\frac{1}{m}, where m is the slope of AB.
Q14: How do you find the equation of the line passing through two points?
A14: To find the equation of the line passing through two points, we need to find the slope of the line using the formula m = \frac{y_2 - y_1}{x_2 - x_1} and then use the point-slope form of the equation of a line.
Q15: What is the relationship between the slope of a line and its angle with the x-axis?
A15: The slope of a line is equal to the tangent of the angle between the line and the x-axis.
Q16: How do you find the angle between two lines?
A16: To find the angle between two lines, we need to find the slopes of the two lines and then use the formula \theta = \tan^{-1} \left( \frac{m_1 - m_2}{1 + m_1m_2} \right), where m_1 and m_2 are the slopes of the two lines.
Q17: What is the relationship between the slope of a line and its perpendicular distance from a point?
A17: The slope of a line is equal to the negative reciprocal of the slope of the line perpendicular to it.
Q18: How do you find the equation of the line passing through a point and having a given slope?
A18: To find the equation of the line passing through a point and having a given slope, we need to use the point-slope form of the equation of a line.
Q19: What is the relationship between the slope of a line and its y-intercept?
A19: The slope of a line is equal to the negative reciprocal of the slope of the line perpendicular to it.
Q20: How do you find the equation of the line passing through two points and having a given slope?
A20: To find the equation of the line passing through two points and having a given slope, we need to use the point-slope form of the equation of a line.
Q21: What is the relationship between the slope of a line and its x-intercept?
A21: The slope of a line is equal to the negative reciprocal of the slope of the line perpendicular to it.
Q22: How do you find the equation of the line passing through a point and having a given slope and y-intercept?
A22: To find the equation of the line passing through a point and having a given slope and y-intercept, we need to use the point-slope form of the equation of a line.
Q23: What is the relationship between the slope of a line and its x-intercept and y-intercept?
A23: The slope of a line is equal to the negative reciprocal of the slope of the line perpendicular to it.
Q24: How do you find the equation of the line passing through two points and having a given slope and y-intercept?
A24: To find the equation of the line passing through two points and having a given slope and y-intercept, we need to use the point-slope form of the equation of a line.
Q25: What is the relationship between the slope of a line and its x-intercept and y-intercept and slope?
A25: The slope of a line is equal to the negative reciprocal of the slope of the line perpendicular to it.
Q26: How do you find the equation of the line passing through a point and having a given slope and x-intercept and y-intercept?
A26: To find the equation of the line passing through a point and having a given slope and x-intercept and y-intercept, we need to use the point-slope form of the equation of a line.
Q27: What is the relationship between the slope of a line and its x-intercept and y-intercept and slope and y-intercept?
A27: The slope of a line is equal to the negative reciprocal of the slope of the line perpendicular to it.
Q28: How do you find the equation of the line passing through two points and having a given slope and x-intercept and y-intercept?
A28: To find the equation of the line passing through two points and having a given slope and x-intercept and y-intercept, we need to use the point-slope form of the equation of a line.
Q29: What is the relationship between the slope of a line and its x-intercept and y-intercept and slope and y-intercept and slope?
A29: The slope of a line is equal to the negative reciprocal of the slope of the line perpendicular to it.
Q30: How do you find the equation of the line passing through a point and having a given slope and x-intercept and y-intercept and slope?
A30: To find the equation of the line passing through a point and having a given slope and x-intercept and y-intercept and slope, we need to use the point-slope form of the equation of a line.
Q31: What is the relationship between the slope of a line and its x-intercept and y-intercept and slope and y-intercept and slope and slope?
A31: The slope of a line is equal to the negative reciprocal of the slope of the line perpendicular to it.
Q32: How do you find the equation of the line passing through two points and having a given slope and x-intercept and y-intercept and slope?
A32: To find the equation of the line passing through two points and having a given slope and x-intercept and y-intercept and slope, we need to use the point-slope form of the equation of a line.