Problem 54
Question
Let \(L\) denote the line in the \(x y\)-plane with \(x\) and \(y\) intercepts as 3 and 1 respectively. Then the image of the point \((-1,-4)\) in this line is: (a) \(\left(\frac{11}{5}, \frac{28}{5}\right)\) (b) \(\left(\frac{29}{5}, \frac{8}{5}\right)\) (c) \(\left(\frac{8}{5}, \frac{29}{5}\right)\) (d) \(\left(\frac{29}{5}, \frac{11}{5}\right)\)
Step-by-Step Solution
Verified Answer
The image of point \((-1,-4)\) is \(\left(\frac{11}{5}, \frac{28}{5}\right)\) (option a).
1Step 1: Identify the Equation of the Line
To determine the equation of the line with intercepts at \( x = 3 \) and \( y = 1 \), we use the intercept form \( \frac{x}{a} + \frac{y}{b} = 1 \), where \( a = 3 \) and \( b = 1 \). Thus, the equation of the line is \( \frac{x}{3} + \frac{y}{1} = 1 \) or \( x + 3y = 3 \).
2Step 2: Calculate Slope of Line Perpendicular to L
The slope of the given line \( L \) is \( -\frac{1}{3} \). The slope of the perpendicular line is the negative reciprocal, which is 3.
3Step 3: Create Equation of Perpendicular Line
Since the line perpendicular to \( L \) passes through \((-1,-4)\) and has slope 3, we use the point-slope form: \( y + 4 = 3(x + 1) \). Simplifying this equation gives \( y = 3x - 1 \).
4Step 4: Find Intersection Point
To find the intersection of \( x + 3y = 3 \) and \( y = 3x - 1 \), substitute \( y = 3x - 1 \) into the first equation: \( x + 3(3x - 1) = 3 \). Simplify to get \( 10x - 3 = 3 \), leading to \( x = \frac{3}{5} \). Substitute back into \( y = 3(\frac{3}{5}) - 1 \), yielding \( y = \frac{4}{5} \). Thus, the intersection is at \( (\frac{3}{5}, \frac{4}{5}) \).
5Step 5: Calculate Image Point
Using the midpoint formula, where \((-1,-4)\) and the intersection \((\frac{3}{5}, \frac{4}{5})\) is a midpoint between the original and image points, set up the system: \( \frac{x + (-1)}{2} = \frac{3}{5} \) and \( \frac{y + (-4)}{2} = \frac{4}{5} \). Solve these to find \( x \) and \( y \). Simplifying, you find \( x = \frac{11}{5} \) and \( y = \frac{28}{5} \).
6Step 6: Verify with Options
The calculated image point \( \left(\frac{11}{5}, \frac{28}{5}\right)\) matches option (a), confirming it is the correct answer.
Key Concepts
Line EquationSlopeIntersection PointPerpendicular Lines
Line Equation
In coordinate geometry, a line equation represents the relationship between the x and y coordinates of any point on the line. For a line in the xy-plane, if the x and y intercepts are given, we use the intercept form to find its equation. This form is represented as \( \frac{x}{a} + \frac{y}{b} = 1 \), where \( a \) and \( b \) are the x-intercept and y-intercept, respectively.
For the line \( L \) in the given exercise, the x-intercept is 3 and the y-intercept is 1. Substituting these values into the intercept form gives \( \frac{x}{3} + \frac{y}{1} = 1 \).
This rearranges to the standard form of \( x + 3y = 3 \). This equation tells us that any point \((x, y)\) lying on this line satisfies this relationship.
For the line \( L \) in the given exercise, the x-intercept is 3 and the y-intercept is 1. Substituting these values into the intercept form gives \( \frac{x}{3} + \frac{y}{1} = 1 \).
This rearranges to the standard form of \( x + 3y = 3 \). This equation tells us that any point \((x, y)\) lying on this line satisfies this relationship.
Slope
The slope of a line in the coordinate plane is a measure of its steepness. It is defined as the ratio of the change in the y-coordinate to the change in the x-coordinate between any two points on the line. Mathematically, it can be given as \( m = \frac{\Delta y}{\Delta x} \).
In the equation \( x + 3y = 3 \), rewriting it in slope-intercept form \( y = mx + c \), we find it is \( y = -\frac{1}{3}x + 1 \). This shows that the slope \( m \) is \(-\frac{1}{3}\).
The slope tells us that for every unit increase in \( x \), \( y \) decreases by \( \frac{1}{3} \) units, indicating it's a downward-slopping line.
In the equation \( x + 3y = 3 \), rewriting it in slope-intercept form \( y = mx + c \), we find it is \( y = -\frac{1}{3}x + 1 \). This shows that the slope \( m \) is \(-\frac{1}{3}\).
The slope tells us that for every unit increase in \( x \), \( y \) decreases by \( \frac{1}{3} \) units, indicating it's a downward-slopping line.
Intersection Point
In coordinate geometry, the intersection point of two lines is the point where they meet or cross each other. To find this point, solve the system of equations formed by the lines.
Given the equations \( x + 3y = 3 \) and \( y = 3x - 1 \), substitute \( y = 3x - 1 \) into the first equation:
\( x + 3(3x - 1) = 3 \).
This simplifies to \( 10x - 3 = 3 \), resulting in \( x = \frac{3}{5} \).
By substituting \( x = \frac{3}{5} \) back into \( y = 3x - 1 \), we find \( y = \frac{4}{5} \).
Therefore, the intersection point is \( (\frac{3}{5}, \frac{4}{5}) \). This point is significant as it helps determine the image point of a reflection across line \( L \).
Given the equations \( x + 3y = 3 \) and \( y = 3x - 1 \), substitute \( y = 3x - 1 \) into the first equation:
\( x + 3(3x - 1) = 3 \).
This simplifies to \( 10x - 3 = 3 \), resulting in \( x = \frac{3}{5} \).
By substituting \( x = \frac{3}{5} \) back into \( y = 3x - 1 \), we find \( y = \frac{4}{5} \).
Therefore, the intersection point is \( (\frac{3}{5}, \frac{4}{5}) \). This point is significant as it helps determine the image point of a reflection across line \( L \).
Perpendicular Lines
Perpendicular lines intersect at right angles (90 degrees). If a line has a slope of \( m \), a line perpendicular to it will have a slope that is the negative reciprocal of \( m \).
In our exercise, the line \( L \) has a slope of \(-\frac{1}{3}\).
Therefore, any line perpendicular to \( L \) would have a slope of 3 (the negative reciprocal of \(-\frac{1}{3}\)).
The perpendicular line in this exercise passes through the point \((-1, -4)\) and has a slope of 3. Using the point-slope form \( y - y_1 = m(x - x_1) \), it becomes \( y + 4 = 3(x + 1) \), which simplifies to \( y = 3x - 1 \).
This understanding of perpendicular lines is important for solving problems involving reflections, as it helps determine the correct orientation of the reflected point across line \( L \).
In our exercise, the line \( L \) has a slope of \(-\frac{1}{3}\).
Therefore, any line perpendicular to \( L \) would have a slope of 3 (the negative reciprocal of \(-\frac{1}{3}\)).
The perpendicular line in this exercise passes through the point \((-1, -4)\) and has a slope of 3. Using the point-slope form \( y - y_1 = m(x - x_1) \), it becomes \( y + 4 = 3(x + 1) \), which simplifies to \( y = 3x - 1 \).
This understanding of perpendicular lines is important for solving problems involving reflections, as it helps determine the correct orientation of the reflected point across line \( L \).
Other exercises in this chapter
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