{"id":281,"date":"2022-04-13T10:33:28","date_gmt":"2022-04-13T10:33:28","guid":{"rendered":"http:\/\/10thclass.deltapublications.in\/?page_id=281"},"modified":"2025-03-01T09:20:08","modified_gmt":"2025-03-01T09:20:08","slug":"n-10-dilations-and-parallel-lines","status":"publish","type":"page","link":"https:\/\/10thclass.deltapublications.in\/index.php\/n-10-dilations-and-parallel-lines\/","title":{"rendered":"N.10 Dilations and parallel lines"},"content":{"rendered":"\n

Dilations and parallel lines<\/strong><\/h2>\n\n\n\n

Key Notes :<\/p>\n\n\n\n

The image of a line after a dilation is also a line.<\/p>\n\n\n\n

The image of the point (x , y) dilated with a scale factor of s centered at the origin is (sx , sy).<\/p>\n\n\n\n

Learn with an example<\/strong><\/p>\n\n\n\n

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Line \ud835\udcc1 has the equation y=-2x-4. Write the equation of the image of \ud835\udcc1 after a dilation with a scale factor of 1\/4, centred at the origin.<\/strong><\/p>\n\n\n\n

Write your answer in slope-intercept form.<\/p>\n\n\n\n

Y = _________<\/p>\n<\/div><\/div>\n\n\n\n

You want to find the equation of the image of \ud835\udcc1 after a dilation with a scale factor of 1\/4 , centred at the origin. Call this image \ud835\udcc1’<\/p>\n\n\n\n

To find the equation of \ud835\udcc1’ , you need two points that lie on  \ud835\udcc1’ .Begin by finding two points that lie on \ud835\udcc1 .<\/p>\n\n\n\n

Start with the y-intercept. Since the equation of \ud835\udcc1 in slope-intercept form is y=-2x-4 the y-intercept is (0 , -4).<\/p>\n\n\n\n

Next, since the slope of \ud835\udcc1 is -2 , which can be written as -2\/1 , move down 2 and right 1 from (0 , -4) to find a second point on \ud835\udcc1 , (1 , -6).<\/p>\n\n\n\n

So, the points  (0 , -4) and (1 , -6) lie on \ud835\udcc1  To find two points on \ud835\udcc1‘<\/em>, apply the dilation<\/p>\n\n\n\n

(X , Y) \u21a6 (1\/4 X , 1\/4 Y )<\/p>\n\n\n\n

(0 , -4) \u21a6 (0 , -4\/4 ) = (0 ,-1)<\/p>\n\n\n\n

(1 , -6) \u21a6 (1\/4 , -6\/4 ) = (1\/4 , -3\/2)<\/p>\n\n\n\n

The image of the y -intercept of \ud835\udcc1 is (0 ,-1) , which is the  y-intercept of \ud835\udcc1’ . In general, the y-intercept of  a line’s image after a dilation centred at the origin is the image of the y-intercept  of the original line. This is because the x-coordinate of the y-intercept is 0, so multiplying by the scale factor of the dilation does not change its value.<\/p>\n\n\n\n

Next, use the slope formula to find the slope of \ud835\udcc1‘<\/em>.<\/p>\n\n\n\n

Slope of \ud835\udcc1‘<\/em> = Y2<\/sub> -Y1<\/sub> \/ X2<\/sub> -X1<\/sub> Slope formula<\/p>\n\n\n\n

= -3\/2- -1 \/ 1\/4-0 Plug in Y2<\/sub> = -3\/2 , Y1<\/sub> = -1 ,X2<\/sub> = 1\/4 and  X1<\/sub> =0<\/p>\n\n\n\n

= -1\/2 \/ 1\/4 Subtract<\/p>\n\n\n\n

= -1\/2 . 4\/1 To divide, multiply by the reciprocal<\/p>\n\n\n\n

= -4\/2 Multiply<\/p>\n\n\n\n

= -2 Simplify<\/p>\n\n\n\n

So, the slope of \ud835\udcc1‘<\/em>  is -2 which is the same as the slope of \ud835\udcc1 . Since \ud835\udcc1‘<\/em> and \ud835\udcc1 have the same slope but different y-intercepts , they are parallel. In general, if a line does not pass through the centre of the dilation, then it is parallel to its image.<\/p>\n\n\n\n

Finally, since \ud835\udcc1‘<\/em> has a slope of -2 and a y-intercept of – 1, the equation of \ud835\udcc1‘<\/em> in slope-intercept form is Y = -2X-1.<\/p>\n<\/div><\/div>\n\n\n\n

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Line \ud835\udcc1 has the equation y=1\/3x+3. Write the equation of the image of \ud835\udcc1 after a dilation with a scale factor of 2, centred at the origin.<\/strong><\/p>\n\n\n\n

Write your answer in slope-intercept form.<\/p>\n\n\n\n

Y = _________<\/p>\n<\/div><\/div>\n\n\n\n

You want to find the equation of the image of \ud835\udcc1 after a dilation with a scale factor of 2 , centred at the origin. Call this image \ud835\udcc1’<\/p>\n\n\n\n

To find the equation of \ud835\udcc1’ , you need two points that lie on  \ud835\udcc1’ .Begin by finding two points that lie on \ud835\udcc1 .<\/p>\n\n\n\n

Start with the y-intercept. Since the equation of \ud835\udcc1 in slope-intercept form is y=1\/3x+3 the y-intercept is (0 , 3).<\/p>\n\n\n\n

Next, since the slope of \ud835\udcc1 is 1\/3 , move up  1 and right 3 from (0 , 3) to find a second point on \ud835\udcc1 , (3 , 4).<\/p>\n\n\n\n

So, the points  (0 , 3) and (3 , 4) lie on \ud835\udcc1  To find two points on \ud835\udcc1‘<\/em>, apply the dilation<\/p>\n\n\n\n

(X , Y) \u21a6 (2 X , 2 Y )<\/p>\n\n\n\n

(0 , 3) \u21a6 (0 , 6 )<\/p>\n\n\n\n

(3 , 4) \u21a6 (6 , 8 ) <\/p>\n\n\n\n

The image of the y -intercept of \ud835\udcc1 is (0 ,6) , which is the  y-intercept of \ud835\udcc1’ . In general, the y-intercept of  a line’s image after a dilation centred at the origin is the image of the y-intercept  of the original line. This is because the x-coordinate of the y-intercept is 0, so multiplying by the scale factor of the dilation does not change its value.<\/p>\n\n\n\n

Next, use the slope formula to find the slope of \ud835\udcc1‘<\/em>.<\/p>\n\n\n\n

Slope of \ud835\udcc1‘<\/em> = Y2<\/sub> -Y1<\/sub> \/ X2<\/sub> -X1<\/sub> Slope formula<\/p>\n\n\n\n

= 8-6 \/ 6-0 Plug in Y2<\/sub> =8 , Y1<\/sub> = 6 ,X2<\/sub> = 6 and  X1<\/sub> =0<\/p>\n\n\n\n

=2\/6 Subtract<\/p>\n\n\n\n

= -1\/2 . 4\/1 To divide, multiply by the reciprocal<\/p>\n\n\n\n

= 1\/3 Simplify<\/p>\n\n\n\n

So, the slope of \ud835\udcc1‘<\/em>  is 1\/3 which is the same as the slope of \ud835\udcc1 . Since \ud835\udcc1‘<\/em> and \ud835\udcc1 have the same slope but different y-intercepts , they are parallel. In general, if a line does not pass through the centre of the dilation, then it is parallel to its image.<\/p>\n\n\n\n

Finally, since \ud835\udcc1‘<\/em> has a slope of 1\/3 and a y-intercept of 6, the equation of \ud835\udcc1‘<\/em> in slope-intercept form is Y = 1\/3 X +6.<\/p>\n<\/div><\/div>\n\n\n\n

let’s practice!<\/p>\n\n\n\n

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\"\"<\/a><\/figure>\n<\/div>\n\n\n\n
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\"\"<\/figure>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Dilations and parallel lines Key Notes : The image of a line after a dilation is also a line. The image of the point (x , y) dilated with a scale factor of s centered at the origin is (sx , sy). Learn with an example Line \ud835\udcc1 has the equation y=-2x-4. Write the equation of the image of \ud835\udcc1 after a dilation with a scale factor of 1\/4, centredContinue reading “N.10 Dilations and parallel lines”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-281","page","type-page","status-publish","hentry","entry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/281","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/comments?post=281"}],"version-history":[{"count":13,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/281\/revisions"}],"predecessor-version":[{"id":17552,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/281\/revisions\/17552"}],"wp:attachment":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/media?parent=281"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}