Question

The midpoint of the line joining the points P (3, 3) and Q (3, -3) is .

• A. (3 , 0)
• B. (3 , 6)
• C. $(3,\frac{3}{2})$

Answer 1

The correct option is A (3 , 0)


Let A(x,y) be the midpoint of line joining P (3, 3) and Q (3, -3). Then the distance of midpoint from both P and Q should be equal, and as midpoint lies in the same line, its abscissa must be 3.

By distance formula,
$(x-3{)}^2+(y-3{)}^2)=(x-3{)}^2+(y-(-3){)}^2$
$(x-3{)}^2+(y-3{)}^2)=(x-3{)}^2+(y+3{)}^2$
$\Rightarrow 12y=0$
$\Rightarrow y=0$

$\therefore$ The midpoint of the line joining the points P (3, 3) and Q (3, -3) is (3,0).