question_answer

If (- 2, 6) is the image of the point (4, 2) with respect to line L = 0, then L =                                      [EAMCET 2002]

A)            3x ? 2y + 5                                 

B)            3x ? 2y + 10

C)            2x + 3y ? 5                                 

D)            6x ? 4y ? 7

Correct Answer: A

Solution :

           The midpoint of \[P(-2,\,6)\] and \[Q(4,\,2)\] is \[\left( \frac{-2+4}{2},\,\frac{6+2}{2} \right)\]  i.e., \[(1,\,4)\] and the gradient of line                          \[PQ=\frac{2-6}{4+2}=\frac{-2}{3}\]                    \ The slope of \[L=\frac{3}{2}\]                    Hence the equation of line which passes through point \[(1,\,4)\] is \[y-4=\frac{3}{2}(x-1)\] Þ \[3x-2y+5=0\] .