{"id":3503,"date":"2021-07-28T12:29:26","date_gmt":"2021-07-28T12:29:26","guid":{"rendered":"https:\/\/mathemerize.com\/?p=3503"},"modified":"2021-11-30T16:44:46","modified_gmt":"2021-11-30T11:14:46","slug":"how-to-separate-pair-of-straight-lines","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/how-to-separate-pair-of-straight-lines\/","title":{"rendered":"How to Separate Pair of Straight Lines"},"content":{"rendered":"

In this post, you will learn how to separate pair of straight lines with example.<\/p>\n

Let’s begin –<\/p>\n

How to Separate Pair of Straight Lines<\/h2>\n

(i) Let us consider the homogeneous equation of second degree as<\/p>\n

\n

\\(ax^2+2hxy+by^2\\) = 0  ……(i)<\/p>\n<\/blockquote>\n

which represent pair of straight lines passes through the origin.<\/p>\n

Now, we divide by \\(x^2\\), we get<\/p>\n

\n

a + \\(2h({y\\over x})\\) + \\(b{({y\\over x})}^2\\) = 0<\/p>\n

Let  \\(y\\over x\\) = m  (say)<\/p>\n

then a + 2hm + \\(bm^2\\) = 0  …….(ii)<\/p>\n<\/blockquote>\n

If \\(m_1\\) & \\(m_2\\) are the roots of equation (ii),<\/p>\n

\n

then \\(m_1\\) + \\(m_2\\) = -\\(2h\\over b\\), \\(m_1m_2\\) = \\(a\\over b\\)<\/p>\n

and also tan\\(\\theta\\) = \\(\\pm\\)\\(2\\sqrt{h^2-ab}\\over {a+b}\\)<\/p>\n<\/blockquote>\n

These lines will be :<\/p>\n

\n

(1) Real and different, if \\(h^2-ab\\) > 0<\/p>\n

(2) Real and Coincident, if \\(h^2-ab\\) = 0<\/p>\n

(3) Imaginary if \\(h^2-ab\\) < 0<\/p>\n<\/blockquote>\n

(ii) The condition that these lines are :<\/p>\n

\n

(1) At Right angles to each other, is a + b = 0<\/p>\n

(2) Coincident is \\(h^2\\) = ab<\/p>\n

(3) Equally inclined to the axes of x is h = 0. i.e. coefficient of xy = 0.<\/p>\n<\/blockquote>\n

(iii) Homogeneous equation of second degree \\(ax^2+2hxy+by^2\\) = 0<\/strong> always represent a pair of straight lines whose equations are<\/p>\n

\n

y = (\\(-h \\pm \\sqrt{h^2-ab}\\over b\\))x = y = \\(m_1\\)x & y = \\(m_2\\)x and \\(m_1\\) + \\(m_2\\) = -\\(2h\\over b\\) ; \\(m_1m_2\\) = \\(a\\over b\\)<\/p>\n<\/blockquote>\n

These straight lines passes through the origin<\/p>\n

(iv) Pair of straight lines perpendicular to the line \\(ax^2+2hxy+by^2\\) = 0<\/strong> and through the origin are given by<\/p>\n

\n

\\(bx^2-2hxy+ay^2\\) = 0<\/p>\n<\/blockquote>\n

(v) The Product of the perpendiculars drawn from the point (\\(x_1,y_1\\)) on the lines \\(ax^2+2hxy+by^2\\) = 0 <\/strong>is<\/p>\n

\n

|\\(a{x_1}^2 + 2hx_1y_1 + b{y_1}^2\\over {\\sqrt{{(a-b)}^2 + 4h^2}}\\)|<\/p>\n<\/blockquote>\n

Note : A homogeneous equation of degree n represent n straight lines passing through origin.<\/p>\n

General Equation and Homogeneous Equation of second degree :<\/strong><\/p>\n

(i)  The general equation of second degree \\(ax^2+2hxy+by^2+2gx+2fy+c\\) = 0 represents a pair of straight lines, if<\/p>\n

\n

\\(\\Delta\\) = \\(abc+2fgh-af^2-bg^2-ch^2\\) = 0<\/p>\n<\/blockquote>\n

(ii)  The product of the perpendiculars drawn from origin to the lines \\(ax^2+2hxy+by^2+2gx+2fy+c\\) = 0 is<\/p>\n

\n

|\\(c\\over \\sqrt{{(a-b)}^2 + 4h^2}\\)|<\/p>\n<\/blockquote>\n\n\n

Example : <\/span> Find the separate equation of the following pair of straight lines \\(x^2 + 4xy + y^2\\) = 0<\/p>\n

Solution : <\/span>Divide the given equation by \\(x^2\\)

\n\\(1 + 4{y\\over x} + {y^2\\over x^2}\\) = 0

\nLet y\/x = m

\n\\(\\implies\\) \\(1 + 4m + m^2\\) = 0 \\(\\implies\\) h = 2 and a = 1 and b = 1

\nLet \\(m_1\\) & \\(m_2\\) are the roots of equation , then \\(m_1\\) + \\(m_2\\) = -\\(2h\\over b\\) = -4 , \\(m_1m_2\\) = \\(a\\over b\\) = 1

\n\\(\\implies\\) \\(m_1\\) = \\(-2 + \\sqrt{3}\\) and \\(m_2\\) = \\(-2 – \\sqrt{3}\\)

\nHence separate pair of straight lines are y = \\(m_1\\)x and y = \\(m_2\\)x

\n\\(\\implies\\) y = (\\(-2 + \\sqrt{3}\\))x and y = (\\(-2 – \\sqrt{3}\\))x

<\/p>\n\n\n

Hope you learnt how to separate pair of straight lines, learn more concepts of straight lines and practice more questions to get ahead in the competition. Good luck!<\/p>\n\n\n

\n
Next – Family of Lines \u2013 The combined Equation of Angle Bisectors<\/a><\/div>\n<\/div>\n\n\n\n
\n
Previous – What is the Formula for Triangle Area ?<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

In this post, you will learn how to separate pair of straight lines with example. Let’s begin – How to Separate Pair of Straight Lines (i) Let us consider the homogeneous equation of second degree as \\(ax^2+2hxy+by^2\\) = 0  ……(i) which represent pair of straight lines passes through the origin. Now, we divide by \\(x^2\\), …<\/p>\n

How to Separate Pair of Straight Lines<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","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":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[32],"tags":[657,656],"yoast_head":"\nHow to Separate Pair of Straight Lines - Mathemerize<\/title>\n<meta name=\"description\" content=\"In this post, you will learn how to separate pair of straight lines from the equation of straight lines with examples.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/how-to-separate-pair-of-straight-lines\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"How to Separate Pair of Straight Lines - 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