What is the equation of the line 2x + 3y = -6 after a dilation of 1/2 centered at the origin?
x-intercept of original: let y = 0, x = -3
y-intercept of original: let x = 0, y = -2
So the new y-intercept must be -1
and of course the slope does not change
we would have
2x + 3y = c
sub in (0,-1)
0 - 3 = c
the new equation is 2x + 3y = -3
You start with 2x + 3y = -6
after applying the dilation of 1/2, all the x- and y- values are 1/2 as big.
So, using the new values,
2(x/2) + 3(y/2) = 1/2 (2x+3y) = 1/2(-6)
So now,
2x+3y = -3
To find the equation of a line after a dilation, we need to apply the scale factor to both the x and y-coordinates. In this case, the scale factor is 1/2.
The equation of the line is 2x + 3y = -6.
To dilate the line by a scale factor of 1/2, we multiply both the x and y-coefficients by 1/2.
(1/2) * 2x + (1/2) * 3y = (1/2) * -6
Simplifying the equation, we get:
x + (3/2)y = -3
Therefore, the equation of the line after a dilation of 1/2 centered at the origin is x + (3/2)y = -3.
To find the equation of the line after a dilation of 1/2 centered at the origin, we need to apply the dilation to the given equation.
A dilation centered at the origin multiplies the x and y coordinates of each point by the dilation factor. In this case, the dilation factor is 1/2, which means we will multiply both x and y by 1/2.
Let's start with the original equation: 2x + 3y = -6
To apply the dilation, we multiply both sides of the equation by 1/2:
(1/2)(2x + 3y) = (1/2)(-6)
Simplifying the left side of the equation:
x + (3/2)y = -3
So, the equation of the line after a dilation of 1/2 centered at the origin is x + (3/2)y = -3.