For new line, slope m=2 and y-intercept c=(-10)
Step-by-step explanation:
Note : Figure given is for reference to understand better.
Where redline is for given line and blueline for new line
The equation of given line 8x-4y=5 and it is dilated by a scale factor of 8 centered at the origin.
Step 1 : Find two points on given line.
When x=0, y=?
[tex]8x-4y=5[/tex]
[tex]8(0)-4y=5[/tex]
[tex]y=\frac{-5}{4}[/tex]
When y=0, x=?
[tex]8x-4y=5[/tex]
[tex]8x-4(0)=5[/tex]
[tex]x=\frac{5}{8}[/tex]
We get points [tex]A(0,\frac{-5}{4}), B(\frac{5}{8},0)[/tex]
Step 2: Find distance from centered and scale it.
Now, It is said that line 8x-4y=5 dilated by a scale factor of 8 centered at the origin and point A and point B is on same.
So that point A and point B will also get dilated by a scale factor of 8 centered at the origin or distance of points from origin will be scaled by 8.
For point A:
Distance of point [tex]A(0,\frac{-5}{4})[/tex] from origin is [tex]( \frac{-5}{4})[/tex] unit in x-direction and zero [tex]\frac{-5}{4})[/tex] unit in y-direction.
After scaled by factor of 8, the distance will multipy by 8 and new location is [tex]A'(0,-10)[/tex]
For point B:
Distance of point [tex]B(\frac{5}{8},0)[/tex] from origin is [tex](\frac{5}{8})[/tex] unit in x-direction and zero unit in y-direction.
After scaled by factor of 8, the distance will multipy by 8 and new location is [tex]B'(5,0)[/tex]
Step 3: Find Equation of new line.
Points [tex]A'(0,-10)[/tex] and [tex]B'(5,0)[/tex] make a new line
The equation of given as
[tex]\frac{y-Y1}{x-X1} = \frac{Y2-Y1}{X2-X1}[/tex]
[tex]\frac{y-(-10)}{x-0} = \frac{0-(-10)}{5-0}[/tex]
[tex]\frac{y+(10)}{x} = 2[/tex]
[tex]\frac{y+(10)}{x} = 2[/tex]
[tex]y+10= 2x[/tex]
[tex]y= 2x-10[/tex]
Now, Comparing with the equation of the line : y=mx + c
Where m=slope and c is the y-intercept
We get, Slope m=2 and y-intercept c=(-10)
