{"id":249073,"date":"2025-07-08T22:08:20","date_gmt":"2025-07-08T22:08:20","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=249073"},"modified":"2025-07-19T06:36:07","modified_gmt":"2025-07-19T06:36:07","slug":"find-the-equation-of-the-line-that-is-perpendicular-to-the-graph-of-3x-2y-12-and-contains-the-point-2-3-then-write-the-equation-of-the-line-in-slope-intercept-form","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/08\/find-the-equation-of-the-line-that-is-perpendicular-to-the-graph-of-3x-2y-12-and-contains-the-point-2-3-then-write-the-equation-of-the-line-in-slope-intercept-form\/","title":{"rendered":"Find the equation of the line that is perpendicular to the graph of 3x + 2y = 12 and contains the point (2, -3), then write the equation of the line in slope-intercept form."},"content":{"rendered":"\n

Find the equation of the line that is perpendicular to the graph of 3x + 2y = 12 and contains the point (2, -3), then write the equation of the line in slope-intercept form.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To find the equation of the line that is perpendicular to the graph of 3x+2y=123x + 2y = 123x+2y=12 and contains the point (2,\u22123)(2, -3)(2,\u22123), we follow these steps:<\/p>\n\n\n\n

Step 1: Find the slope of the given line<\/h3>\n\n\n\n

First, we need to rewrite the equation 3x+2y=123x + 2y = 123x+2y=12 in slope-intercept form<\/strong>, which is y=mx+by = mx + by=mx+b, where mmm is the slope of the line.<\/p>\n\n\n\n

Start by isolating yyy in the equation: 3x+2y=123x + 2y = 123x+2y=12<\/p>\n\n\n\n

Subtract 3x3x3x from both sides: 2y=\u22123x+122y = -3x + 122y=\u22123x+12<\/p>\n\n\n\n

Now, divide both sides by 2: y=\u221232x+6y = -\\frac{3}{2}x + 6y=\u221223\u200bx+6<\/p>\n\n\n\n

The slope of the given line is m=\u221232m = -\\frac{3}{2}m=\u221223\u200b.<\/p>\n\n\n\n

Step 2: Find the slope of the perpendicular line<\/h3>\n\n\n\n

The slope of a line that is perpendicular<\/strong> to another is the negative reciprocal<\/strong> of the original slope. The negative reciprocal of \u221232-\\frac{3}{2}\u221223\u200b is 23\\frac{2}{3}32\u200b.<\/p>\n\n\n\n

So, the slope of the perpendicular line is 23\\frac{2}{3}32\u200b.<\/p>\n\n\n\n

Step 3: Use the point-slope form to write the equation<\/h3>\n\n\n\n

We know the slope of the perpendicular line is 23\\frac{2}{3}32\u200b and that it passes through the point (2,\u22123)(2, -3)(2,\u22123). We can use the point-slope form of the equation of a line: y\u2212y1=m(x\u2212x1)y – y_1 = m(x – x_1)y\u2212y1\u200b=m(x\u2212x1\u200b)<\/p>\n\n\n\n

where m=23m = \\frac{2}{3}m=32\u200b and (x1,y1)=(2,\u22123)(x_1, y_1) = (2, -3)(x1\u200b,y1\u200b)=(2,\u22123).<\/p>\n\n\n\n

Substitute the values: y\u2212(\u22123)=23(x\u22122)y – (-3) = \\frac{2}{3}(x – 2)y\u2212(\u22123)=32\u200b(x\u22122)<\/p>\n\n\n\n

Simplify: y+3=23(x\u22122)y + 3 = \\frac{2}{3}(x – 2)y+3=32\u200b(x\u22122)<\/p>\n\n\n\n

Now, expand the right side: y+3=23x\u221243y + 3 = \\frac{2}{3}x – \\frac{4}{3}y+3=32\u200bx\u221234\u200b<\/p>\n\n\n\n

Step 4: Isolate yyy to get the slope-intercept form<\/h3>\n\n\n\n

Subtract 3 from both sides: y=23x\u221243\u22123y = \\frac{2}{3}x – \\frac{4}{3} – 3y=32\u200bx\u221234\u200b\u22123<\/p>\n\n\n\n

Rewrite 3 as 93\\frac{9}{3}39\u200b: y=23x\u221243\u221293y = \\frac{2}{3}x – \\frac{4}{3} – \\frac{9}{3}y=32\u200bx\u221234\u200b\u221239\u200b<\/p>\n\n\n\n

Combine the constants: y=23x\u2212133y = \\frac{2}{3}x – \\frac{13}{3}y=32\u200bx\u2212313\u200b<\/p>\n\n\n\n

Final Answer:<\/h3>\n\n\n\n

The equation of the line that is perpendicular to 3x+2y=123x + 2y = 123x+2y=12 and contains the point (2,\u22123)(2, -3)(2,\u22123) is: y=23x\u2212133y = \\frac{2}{3}x – \\frac{13}{3}y=32\u200bx\u2212313\u200b<\/p>\n\n\n\n

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

Find the equation of the line that is perpendicular to the graph of 3x + 2y = 12 and contains the point (2, -3), then write the equation of the line in slope-intercept form. The Correct Answer and Explanation is: To find the equation of the line that is perpendicular to the graph of 3x+2y=123x […]<\/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-249073","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/249073","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=249073"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/249073\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=249073"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=249073"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=249073"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}