{"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
(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
\na + \\(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
\nthen \\(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
\ny = (\\(-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\nHope 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