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