{"id":252471,"date":"2025-07-11T10:22:03","date_gmt":"2025-07-11T10:22:03","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=252471"},"modified":"2025-07-11T10:22:05","modified_gmt":"2025-07-11T10:22:05","slug":"points-a-and-b-have-coordinates-of-5-2-and-10-1-respectively","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/11\/points-a-and-b-have-coordinates-of-5-2-and-10-1-respectively\/","title":{"rendered":"Points A and B have coordinates of (5, 2) and (10, -1) respectively."},"content":{"rendered":"\n<p>Points A and B have coordinates of (5, 2) and (10, -1) respectively. What is the equation of the perpendicular bisector of AB.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To find the equation of the perpendicular bisector of the line segment connecting points A(5, 2) and B(10, -1), follow these steps:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Find the midpoint of AB<\/h3>\n\n\n\n<p>The midpoint of a line segment is found by averaging the x-coordinates and the y-coordinates of the two endpoints. Midpoint=(x1+x22,y1+y22)\\text{Midpoint} = \\left( \\frac{x_1 + x_2}{2}, \\frac{y_1 + y_2}{2} \\right)Midpoint=(2&#215;1\u200b+x2\u200b\u200b,2y1\u200b+y2\u200b\u200b)<\/p>\n\n\n\n<p>For points A(5, 2) and B(10, -1): Midpoint=(5+102,2+(\u22121)2)=(152,12)\\text{Midpoint} = \\left( \\frac{5 + 10}{2}, \\frac{2 + (-1)}{2} \\right) = \\left( \\frac{15}{2}, \\frac{1}{2} \\right)Midpoint=(25+10\u200b,22+(\u22121)\u200b)=(215\u200b,21\u200b)<\/p>\n\n\n\n<p>So, the midpoint is (7.5,0.5)\\left( 7.5, 0.5 \\right)(7.5,0.5).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Find the slope of AB<\/h3>\n\n\n\n<p>The slope of a line is given by the formula: slope=y2\u2212y1x2\u2212x1\\text{slope} = \\frac{y_2 &#8211; y_1}{x_2 &#8211; x_1}slope=x2\u200b\u2212x1\u200by2\u200b\u2212y1\u200b\u200b<\/p>\n\n\n\n<p>For points A(5, 2) and B(10, -1): slope&nbsp;of&nbsp;AB=\u22121\u2212210\u22125=\u221235\\text{slope of AB} = \\frac{-1 &#8211; 2}{10 &#8211; 5} = \\frac{-3}{5}slope&nbsp;of&nbsp;AB=10\u22125\u22121\u22122\u200b=5\u22123\u200b<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Find the slope of the perpendicular bisector<\/h3>\n\n\n\n<p>The slope of the perpendicular bisector is the negative reciprocal of the slope of AB. Since the slope of AB is \u22123\/5-3\/5\u22123\/5, the slope of the perpendicular bisector will be 5\/35\/35\/3 (because the negative reciprocal of \u22123\/5-3\/5\u22123\/5 is 5\/35\/35\/3).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Use point-slope form to find the equation of the perpendicular bisector<\/h3>\n\n\n\n<p>The point-slope form of a line equation is: y\u2212y1=m(x\u2212x1)y &#8211; y_1 = m(x &#8211; x_1)y\u2212y1\u200b=m(x\u2212x1\u200b)<\/p>\n\n\n\n<p>where mmm is the slope and (x1,y1)(x_1, y_1)(x1\u200b,y1\u200b) is a point on the line. Using the midpoint (7.5,0.5)(7.5, 0.5)(7.5,0.5) and the slope 5\/35\/35\/3, the equation becomes: y\u22120.5=53(x\u22127.5)y &#8211; 0.5 = \\frac{5}{3}(x &#8211; 7.5)y\u22120.5=35\u200b(x\u22127.5)<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 5: Simplify the equation<\/h3>\n\n\n\n<p>First, distribute 53\\frac{5}{3}35\u200b on the right side: y\u22120.5=53x\u221253\u00d77.5y &#8211; 0.5 = \\frac{5}{3}x &#8211; \\frac{5}{3} \\times 7.5y\u22120.5=35\u200bx\u221235\u200b\u00d77.5 y\u22120.5=53x\u221212.5y &#8211; 0.5 = \\frac{5}{3}x &#8211; 12.5y\u22120.5=35\u200bx\u221212.5<\/p>\n\n\n\n<p>Next, add 0.5 to both sides to isolate yyy: y=53x\u221212.5+0.5y = \\frac{5}{3}x &#8211; 12.5 + 0.5y=35\u200bx\u221212.5+0.5 y=53x\u221212y = \\frac{5}{3}x &#8211; 12y=35\u200bx\u221212<\/p>\n\n\n\n<p>Thus, the equation of the perpendicular bisector of AB is: y=53x\u221212y = \\frac{5}{3}x &#8211; 12y=35\u200bx\u221212<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion<\/h3>\n\n\n\n<p>The equation of the perpendicular bisector of the line segment joining points A(5, 2) and B(10, -1) is y=53x\u221212y = \\frac{5}{3}x &#8211; 12y=35\u200bx\u221212. This line passes through the midpoint of the segment and is perpendicular to it.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-240.jpeg\" alt=\"\" class=\"wp-image-252472\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Points A and B have coordinates of (5, 2) and (10, -1) respectively. What is the equation of the perpendicular bisector of AB. The Correct Answer and Explanation is: To find the equation of the perpendicular bisector of the line segment connecting points A(5, 2) and B(10, -1), follow these steps: Step 1: Find the [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-252471","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/252471","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=252471"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/252471\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=252471"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=252471"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=252471"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}