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STD 11 - 9. straight line — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 9. straight line4 Marks
MCQ
The equation of the line bisecting perpendicularly the segment joining the points $(-4, 6)$ and $(8, 8)$ is
  • ✓
    $6x + y - 19 = 0$
  • B
    $y = 7$
  • C
    $6x + 2y - 19 = 0$
  • D
    $x + 2y - 7 = 0$
✓

Answer

Correct option: A.
$6x + y - 19 = 0$
a
(a) Equation of the line passing through $( - 4,\,6)$ and $(8,\,8)$ is $y - 6 = \left( {\frac{{8 - 6}}{{8 + 4}}} \right)\,(x + 4)$ ==> $y - 6 = \frac{2}{{12}}(x + 4)$

==> $6y - 36 = x + 4$ ==> $6y - x - 40 = 0$ ……$(i)$

Now equation of any line perpendicular to it is

$6x + y + \lambda = 0$ ……$(ii)$

This line passes through the mid point of $( - 4,\,6)$ and $(8,\,8)$ i.e., $(2,\,7)$==> $6 \times 2 + 7 + \lambda = 0$

==> $19 + \lambda = 0 \Rightarrow \lambda = - 19$

From $(ii)$ the equation of required line is $6x + y - 19 = 0$.

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