{"id":2592,"date":"2012-07-31T22:04:30","date_gmt":"2012-07-31T22:04:30","guid":{"rendered":"http:\/\/SchoolTutoring.com\/help\/?p=2592"},"modified":"2014-12-02T08:32:07","modified_gmt":"2014-12-02T08:32:07","slug":"writing-the-equation-of-perpendicular-lines","status":"publish","type":"post","link":"https:\/\/schooltutoring.com\/help\/writing-the-equation-of-perpendicular-lines\/","title":{"rendered":"Writing the Equation of Perpendicular Lines"},"content":{"rendered":"

Two lines are said to be perpendicular if the angle between them is 90\u00ba (90 degrees). There is a result on perpendicular line which states<\/a>: Two lines are perpendicular if and only if the product of their slopes is -1.<\/p>\n

i.e. If the slopes of two lines which are perpendicular are m1<\/strong> and m2<\/strong> then m1 x \u00a0m2=-1.<\/strong><\/p>\n

Example:<\/span><\/strong><\/p>\n

The slope of a line is 3\/2<\/strong>. If a line is perpendicular to this line, what is the slope of this new line?<\/p>\n

Solution:<\/span><\/strong><\/p>\n

By the condition for the slopes of perpendicular lines,<\/p>\n

m1 x \u00a0m2=-1<\/strong><\/p>\n

So, the slope of new line = -2\/3.<\/strong><\/p>\n

Finding the Equation of Perpendicular Lines:<\/h3>\n

For finding the equation of a line perpendicular to ax+by+c=0<\/strong> and passing through the point (x1,y1)<\/strong>, there are 2 methods.<\/p>\n

Method 1:<\/h4>\n

(1)\u00a0\u00a0\u00a0 Finding the slope of ax+by+c=0<\/strong> which can be obtained by the expression \u2013a\/b.<\/strong><\/p>\n

i.e. slope of ax+by+c=0<\/strong> is a\/b<\/strong>.<\/p>\n

(2) \u00a0Finding the slope of the line perpendicular to it.<\/p>\n

i.e. the slope of the perpendicular line, m = b\/a.<\/strong><\/p>\n

(3)\u00a0 Then the corresponding perpendicular line’s equation can be found using point slope form.<\/p>\n

i.e., the required equation is,<\/p>\n

y-y1=m(x-x1)<\/strong><\/p>\n

Example:<\/span><\/strong><\/p>\n

Find the equation of the line perpendicular to 3x+y+3=0<\/strong> and passing through (-1,2).<\/strong><\/p>\n

Solution:<\/span><\/strong><\/em><\/p>\n

The slope of given line, m = -3\/1=-3<\/strong><\/p>\n

The slope of the perpendicular line = 1\/3.<\/strong><\/p>\n

(x1,y1) = (-1,2).<\/strong><\/p>\n

The equation of perpendicular line is,<\/p>\n

y-y1=m(x-x1)<\/strong><\/p>\n

y-2 = 1\/3 (x+1)<\/strong><\/p>\n

3y-6 = x+1<\/strong><\/p>\n

x-3y+7=0.<\/strong><\/p>\n

Method 2:<\/h4>\n

(1)\u00a0\u00a0\u00a0 Take the equation of line perpendicular to ax+by+c=0<\/strong> as bx-ay+k=0<\/strong> where k<\/strong> is a constant.<\/p>\n

(2)\u00a0\u00a0\u00a0 The value of \u2018k<\/strong>\u2019 can be found b substituting the given point (x1,y1)<\/strong> in the equation bx – ay+k=0.<\/strong><\/p>\n

(3)\u00a0\u00a0\u00a0 Substitute \u2018k<\/strong>\u2019 back into the equation bx-ay+k=0<\/strong>, which is the required equation.<\/p>\n

Example:<\/span><\/strong><\/p>\n

Find the equation of the line perpendicular to 3x+y+3=0<\/strong> and passing through (-1,2)<\/strong>.<\/p>\n

Solution:<\/span><\/strong><\/em><\/p>\n

The equation of line perpendicular to 3x+y+3=0<\/strong> is,<\/p>\n

x-3y+k=0.<\/strong><\/p>\n

Here x=-1<\/strong> and y=2<\/strong><\/p>\n

-1-3(2)+k=0<\/strong><\/p>\n

-1-6+k=0<\/strong><\/p>\n

k-7=0<\/strong><\/p>\n

k=7.<\/strong><\/p>\n

So the required equation of the perpendicular line is,<\/p>\n

x-3y+7=0<\/strong><\/p>\n

Two lines are said to be perpendicular if the angle between them is 90\u00ba (90 degrees). There is a result on perpendicular line which states: Two lines are perpendicular if and only if the product of their slopes is -1. i.e. If the slopes of two lines which are perpendicular are m1 and m2 then […]<\/p>\n","protected":false},"author":19,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"inline_featured_image":false,"footnotes":""},"categories":[2],"tags":[588,1319,1396,1662],"class_list":["post-2592","post","type-post","status-publish","format-standard","hentry","category-algebra","tag-equation","tag-perpendicular","tag-product","tag-slopes"],"acf":[],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/posts\/2592","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/users\/19"}],"replies":[{"embeddable":true,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/comments?post=2592"}],"version-history":[{"count":0,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/posts\/2592\/revisions"}],"wp:attachment":[{"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/media?parent=2592"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/categories?post=2592"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/tags?post=2592"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}

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