Question

Please Help Quicly The equation 8x - 4y = 5 is dilated by a scale factor of 8 centered at the origin. What is the new slope and y-intercept after dilation?

Answer 1

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=?

$8x-4y=5$

$8(0)-4y=5$

$y=\frac{-5}{4}$

When y=0, x=?

$8x-4y=5$

$8x-4(0)=5$

$x=\frac{5}{8}$

We get points $A(0,\frac{-5}{4}), B(\frac{5}{8},0)$

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 $A(0,\frac{-5}{4})$ from origin is $( \frac{-5}{4})$  unit in x-direction and zero $\frac{-5}{4})$  unit in y-direction.

After scaled by factor of 8, the distance will multipy by 8 and new location is $A'(0,-10)$

For point B:

Distance of point $B(\frac{5}{8},0)$ from origin is $(\frac{5}{8})$ 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 $B'(5,0)$

Step 3: Find Equation of new line.

Points $A'(0,-10)$ and $B'(5,0)$ make a new line

The equation of given as

$\frac{y-Y1}{x-X1} = \frac{Y2-Y1}{X2-X1}$

$\frac{y-(-10)}{x-0} = \frac{0-(-10)}{5-0}$

$\frac{y+(10)}{x} = 2$

$\frac{y+(10)}{x} = 2$

$y+10= 2x$

$y= 2x-10$

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)