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 m and ll respectively, what is the slope of X'Y' ?

1 Answer
Somebody N.
Jun 2, 2018

color(blue)(m)m

Explanation:

If line xy has end points (x_1,y_1),(x_2,y_2)(x1,y1),(x2,y2)

Dilated line line has endpoints((13x_1)/10,(13y_1)/10),((13x_2)/10,(13y_2)/10)(13x110,13y110),(13x210,13y210)

m=(y_2-y_1)/(x_2-x_1)m=y2−y1x2−x1

After dilation:

m=((13y_2)/10-(13y_1)/10)/((13x_2)/10-(13x_1)/10)=(13/10(y_2-y_1))/(13/10(x_2-x_1))=(y_2-y_1)/(x_2-x_1)m=13y210−13y11013x210−13x110=1310(y2−y1)1310(x2−x1)=y2−y1x2−x1

If:

l=sqrt((x_2-x_1)^2+(y_2-y_1)^2)l=√(x2−x1)2+(y2−y1)2

After dilation:

sqrt(((13x_2)/10-(13x_1)/10)^2+((13y_2)/10-(13y_1)/10)^2)√(13x210−13x110)2+(13y210−13y110)2

sqrt((13/10)^2(x_2-x_1)^2+(13/10)^2(y_2-y_1)^2)√(1310)2(x2−x1)2+(1310)2(y2−y1)2

sqrt((13/10)^2((x_2-x_1)^2+(y_2-y_1)^2))√(1310)2((x2−x1)2+(y2−y1)2)

13/10sqrt(((x_2-x_1)^2+(y_2-y_1)^2))1310√((x2−x1)2+(y2−y1)2)

:.

Dilated length:

13/10l

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.

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