# Line XY is dilated by a scale factor of 1.3 with the origin as the center of dilation to create the image line X'Y' . If the slope and length of XY are  m  and l respectively, what is the slope of X'Y' ?

Jun 2, 2018

$\textcolor{b l u e}{m}$

#### Explanation:

If line xy has end points $\left({x}_{1} , {y}_{1}\right) , \left({x}_{2} , {y}_{2}\right)$

Dilated line line has endpoints$\left(\frac{13 {x}_{1}}{10} , \frac{13 {y}_{1}}{10}\right) , \left(\frac{13 {x}_{2}}{10} , \frac{13 {y}_{2}}{10}\right)$

$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

After dilation:

$m = \frac{\frac{13 {y}_{2}}{10} - \frac{13 {y}_{1}}{10}}{\frac{13 {x}_{2}}{10} - \frac{13 {x}_{1}}{10}} = \frac{\frac{13}{10} \left({y}_{2} - {y}_{1}\right)}{\frac{13}{10} \left({x}_{2} - {x}_{1}\right)} = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

If:

$l = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

After dilation:

$\sqrt{{\left(\frac{13 {x}_{2}}{10} - \frac{13 {x}_{1}}{10}\right)}^{2} + {\left(\frac{13 {y}_{2}}{10} - \frac{13 {y}_{1}}{10}\right)}^{2}}$

$\sqrt{{\left(\frac{13}{10}\right)}^{2} {\left({x}_{2} - {x}_{1}\right)}^{2} + {\left(\frac{13}{10}\right)}^{2} {\left({y}_{2} - {y}_{1}\right)}^{2}}$

$\sqrt{{\left(\frac{13}{10}\right)}^{2} \left({\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}\right)}$

$\frac{13}{10} \sqrt{\left({\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}\right)}$

$\therefore$

Dilated length:

$\frac{13}{10} l$

In the above, the work carried out to find the slope of the dilated line was unnecessary. Dilations do not change the orientation, so the gradient would remain the same.