Question

Question 12
If the point P(2,1) lies on the line segment joining points A(4,2) and B(8,4) then
(A) $AP=\frac{1}{3}AB$
(B) $AP=PB$
(C) $PB=\frac{1}{3}AB$
(D) $AP=\frac{1}{2}AB$

Answer 1

Given that, the point P(2,1) lies on the line segment joining points A(4,2) and B(8,4) which shows in the figure below:

Now, distance between
$A(4,2)andP(2,1),AP=\sqrt{(2-4{)}^2+(1-2{)}^2}\left[\begin{array}{cc}\because \text{Distance between two points}(x_1,y_1)and(x_2,y_2),d& \\ =\sqrt{(x_2-x_1{)}^2+(x_2-y_1{)}^2}& \end{array}\right]=\sqrt{(-2{)}^2+(-1{)}^2}=\sqrt{4+1}=\sqrt{5}\text{Distance between A(4,2) and B(8,4),}AB=\sqrt{(8-4{)}^2+(4-2{)}^2}=\sqrt{(4{)}^2+(2{)}^2}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}$
Distance between B(8,4) and P(2,1), BP
$=\sqrt{(8-2{)}^2+(4-1{)}^2}=\sqrt{6^2+3^2}=\sqrt{36+9}=\sqrt{45}=3\sqrt{5}$
Distance AB = $\sqrt{(4-8{)}^2+(2-4{)}^2}$
= $\sqrt{(16+4)}=16+4=20=2\sqrt{5}$
$\therefore AB=2\sqrt{5}=2AP\Rightarrow AP=\frac{AB}{2}\text{Hence, required condition is AP}=\frac{AB}{2}$