Example 1 : <\/span>Find the equation of lines which passes through the point (3,4) and the sum of intercepts on the axes is 14.<\/p>\n Solution : <\/span>Let the equation of line be \\(x\\over a\\) + \\(y\\over b\\) = 1 …..(i) Example 2 : <\/span> If x + 4y – 5 = 0 and 4x + ky + 7 = 0 are two perpendicular lines then k is – <\/p>\n Solution : <\/span> \\(m_1\\) = -\\(1\\over 4\\) \\(m_2\\) = -\\(4\\over k\\) Example 3 : <\/span> If the straight line 3x + 4y + 5 – k(x + y + 3) = 0 is parallel to y-axis, then the value of k is –<\/p>\n Solution : <\/span>A straight line is parallel to y-axis, if its y-coefficient is zero Example 4 : <\/span> If \\(\\lambda x^2 – 10xy + 12y^2 + 5x – 16y – 3\\) = 0 represents a pair of straight lines, then \\(\\lambda\\) is equal to –<\/p>\n Solution : <\/span>Comparing with \\(ax^2+2hxy+by^2+2gx+2fy+c\\) = 0 Practice these given straight line examples to test your knowledge on concepts of straight lines.<\/p>\n \n <\/div>\n <\/div>\n","protected":false},"excerpt":{"rendered":" Here you will learn some straight line examples for better understanding of straight line concepts. Example 1 : Find the equation of lines which passes through the point (3,4) and the sum of intercepts on the axes is 14. Solution : Let the equation of line be \\(x\\over a\\) + \\(y\\over b\\) = 1 …<\/p>\n
\n This line passes through (3,4), therefore \\(3\\over a\\) + \\(4\\over b\\) = 1 …….(ii)
\n It is given that a + b = 14 \\(\\implies\\) b = 14 – a in (ii), we get
\n \\(3\\over a\\) + \\(4\\over 14 – a\\) = 1 \\(\\implies\\) \\(a^2\\) – 13a + 42 = 0
\n \\(\\implies\\) (a – 7)(a – 6) = 0 \\(\\implies\\) a = 7, 6
\n for a = 7, b = 14 – 7 = 7 and for a = 6, b = 14 – 6 = 8
\n Putting the values of a and b in (i), we get the equations of lines
\n \\(x\\over 7\\) + \\(y\\over 7\\) = 1 and \\(x\\over 6\\) + \\(y\\over 8\\) = 1<\/p>\n
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\n Two lines are perpendicular if \\(m_1 m_2\\) = -1
\n \\(\\implies\\) (-\\(1\\over 4\\))\\(\\times\\)(-\\(4\\over k\\)) = -1 \\(\\implies\\) k = -1\n <\/p>
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\n i.e. 4 – k = 0 i.e. k = 4<\/p>\n
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\n Here a = \\(\\lambda\\), b = 12, c = -3, f = -8, g = 5\/2, h = -5
\n Using condition \\(abc+2fgh-af^2-bg^2-ch^2\\) = 0, we have
\n \\(\\lambda\\)(12)(-3) + 2(-8)(5\/2)(-5) – \\(\\lambda\\)(64) – 12(25\/4) + 3(25) = 0
\n \\(\\implies\\) -36\\(\\lambda\\) + 200 – 64\\(\\lambda\\) – 75 + 75 = 0
\n \\(\\implies\\) 100\\(\\lambda\\) = 200
\n \\(\\therefore\\) \\(\\lambda\\) = 2<\/p>\n
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