Straight Line Examples

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.

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