Step 1 :
Draw the equation of the line given with x and y intercepts.
Step 2 :
Mark the center of enlargement or reduction.
Step 3 :
Step 4 :
Note :
Slope of new line and old line will be equal.
Problem 1 :
Find the equation of the image when
y = 2x is
i) enlargement with center O (0, 0) and scale factor k = 3
ii) reduced with center O (0, 0) and scale factor k = 1/3
Solution :
y = 2x
i) enlargement with scale factor k = 3
(x, y) --> (kx, ky) --> (3x, 3y)
3y = 2(3x)i
3y = 6x
Dividing by 3, we get
y = 2x
ii) reduction with scale factor k = 1/3
(x, y) --> ( kx, ky ) --> (x/3, y/3)
Here the value of k is in between 0 and 1.
y/3 = 2(x/3)
y/3 = 2x/3
y = 2x
Problem 2 :
Find the equation of the image when
y = -x + 2 is
i) enlargement with center O (0, 0) and scale factor k = 4
ii) reduced with center O (0, 0) and scale factor k = 2/3
Solution :
i) Let us find the x and y-intercepts to draw the graph of the line.
x-intercept : Put y = 0 y = -x + 2 -x + 2 = 0 -x = -2 x = 2 |
y-intercept : Put x = 0 y = -x + 2 y = -0 + 2 y = 2 |
(2, 0) and (0, 2) are the x and y-intercepts.
After the enlargement with the scale factor or 4 units :
Slope = 8/8 = 1
It must be a falling line, then m = -1
y-intercept = 8
then,
y = -1x + 8
ii) reduced with center O (0, 0) and scale factor k = 2/3
Distance between center and y-intercept of the old line = 2 units.
Since the reduction to be done with 2/3 units, we have to multiply 2 by 2/3. Then we get 4/3. So, the new y-intercept will be 4/3.
There is no change in slope, then
y = -x + (4/3)
Problem 3 :
Find the equation of the image when
y = 2x + 3 is
i) enlargement with center O (2, 1) and scale factor k = 2
ii) reduced with center O (2, 1) and scale factor k = 1/2
Solution :
i) Let us find the x and y-intercepts to draw the graph of the line.
x-intercept Put y = 0 2x + 3 = 0 2x = -3 x = -3/2 |
y-intercept Put x = 0 2(0) + 3 = y y = 3 |
(-3/2, 0) and (0, 3) are the x and y-intercepts.
Distance between center and point on the old line :
Distance between center and point on the old line :
(-4, 1) and (2, 13)
Equation of new line after enlargement :
y = 2x + b
Applying (-4, 1), we get
1 = 2(-4) + b
1 + 8 = b
b = 9
Applying the value of b, we get
y = 2x + 9
ii) reduced with center O (2, 1) and scale factor k = 1/2
Distance between center and point on the old line :
Distance between center and point on the old line :
(0.5, 1) and (2, 4)
Equation of new line after enlargement :
y = 2x + b
Applying (2, 4), we get
4 = 2(2) + b
4 - 4 = b
b = 0
Applying the value of b, we get
y = 2x + 0
y = 2x
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM