To reflect a point or object in a line, we need to find the perpendicular distance from the point/object to the line, and then move it the same distance on the other side of the line. This will create a mirror image of the point/object with respect to the line.
First, let's find the equation of the perpendicular line passing through the point (0,2) on the line L. Since the slope of L is 2, the slope of the perpendicular line will be -1/2 (negative reciprocal). Using point-slope form, the equation of the perpendicular line is:
y - 2 = (-1/2)(x - 0)
y = (-1/2)x + 2
Next, let's find the point of intersection between the line a: 3x - y - 2 = 0 and the perpendicular line y = (-1/2)x + 2. We can solve this system of equations by substitution:
3x - y - 2 = 0
y = (-1/2)x + 2
Substituting y = (-1/2)x + 2 into the first equation, we get:
3x - (-1/2)x + 4 - 2 = 0
x = 2/7
Substituting x = 2/7 into y = (-1/2)x + 2, we get:
y = (-1/2)(2/7) + 2
y = 17/7
So the point of intersection is (2/7, 17/7).
Next, let's find the perpendicular distance between line a and line L. We can use the formula for the distance between a point (x1, y1) and a line Ax + By + C = 0:
distance = |Ax1 + By1 + C| / sqrt(A^2 + B^2)
For line a: 3x - y - 2 = 0, we have A = 3, B = -1, and C = -2.
distance = |3(2/7) - 17/7 - 2| / sqrt(3^2 + (-1)^2)
distance = 5/√10
Finally, we can reflect the line a by moving the same distance on the other side of line L. Since the slope of line a is positive, the reflected line will be on the other side of line L with the same slope. Using point-slope form and the point of intersection (2/7, 17/7), we get:
y - (17/7) = (1/2)(x - 2/7)
2y - 34/7 = x - 2/7
2y = x + 48/7
y = (1/2)x + 24/7
So the image of line a after reflection in line L:y=2x-5 is y = (1/2)x + 24/7.
b. (x-2)^2 + (y+3)^2 = 4
To reflect a circle in a line, we need to reflect its center and also make sure that the radius remains the same. The center of the given circle is (2,-3) and the radius is 2 (square root of 4).
Let's first reflect the center of the circle in line L. Using the same formula for the perpendicular distance, the distance between the center of the circle and line L is:
distance = |2(2) - (-3) - 5| / sqrt(2^2 + 1^2)
distance = (5/√5) = √5
So the perpendicular distance is √5, and we need to move the center of the circle by 2√5 units (twice the perpendicular distance) on the other side of line L. Since the slope of line L is 2, the slope of the perpendicular line passing through the center of the circle will be -1/2. Using point-slope form, the equation of the perpendicular line passing through (2,-3) is:
y + 3 = (-1/2)(x - 2)
y = (-1/2)x + 2
We need to find the point on this line that is 2√5 units away from (2,-3). Let's call this point P. We can use the distance formula between two points:
distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
2√5 = sqrt((x - 2)^2 + (y + 3)^2)
Squaring both sides, we get:
20 = (x - 2)^2 + (y + 3)^2
Expanding and simplifying, we get:
x^2 - 4x + y^2 + 6y + 5 = 0
This is the equation of a circle centered at (2, -3) with radius sqrt(20) = 2sqrt(5). We need to find the image of this circle after reflection in line L.
The center of the reflected circle will be 2√5 units below the point of intersection between line L and the perpendicular line passing through (2,-3). We can find this point by solving the system of equations:
y = 2x - 5
y = (-1/2)x + 2
By substitution, we get:
2x - 5 = (-1/2)x + 2
5/2 x = 7
x = 14/5
Substituting x = 14/5 into y = 2x - 5, we get:
y = 23/5
So the point of intersection is (14/5, 23/5), and the center of the reflected circle will be:
(14/5, 23/5 - 2√5)
The radius of the reflected circle will be the same as the original circle, which is 2sqrt(5).
Using the standard form of a circle, the equation of the reflected circle is:
(x - 14/5)^2 + (y - (23/5 - 2√5))^2 = (2√5)^2
Simplifying, we get:
(x - 14/5)^2 + (y - (23/5 - 2√5))^2 = 20
So the image of the circle (x-2)^2 + (y+3)^2 = 4 after reflection in line L:y=2x-5 is:
(x - 14/5)^2 + (y - 23/5 + 2√5)^2 = 20.
Find the image of the following lines and circle after reflection in the line
L:y=2x-5
a.3x-y-2=0
1 answer
1 answer