question_answer

If the point P(2,1) lies on the line segment joining points A(4, 2) and 5(8, 4), then ..........

A) \[AP=\frac{1}{3}AB\]

B) \[AP=PB\]

C) \[PB=\frac{1}{3}AB\]

D) \[AP=\frac{1}{2}AB\]

Correct Answer: D

Solution :

 

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

Now, distance between A(4, 2) and P(2, 1),

\[AP=\sqrt{{{\left( 2-4 \right)}^{2}}+{{\left( 1-2 \right)}^{2}}}\]

[\[\because\]distance between two points two points

\[\left( {{x}_{1}},\,{{y}_{1}} \right)\] and \[B\left( {{x}_{2}},\,{{y}_{2}} \right)\]

,           \[\left. d=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}} \right]\]

\[=\sqrt{{{\left( -2 \right)}^{2}}+{{\left( -1 \right)}^{2}}}=\sqrt{4+1}=\sqrt{5}\]units

Distance between A (4, 2) and B (8, 4),

\[AB=\sqrt{{{\left( 8-4 \right)}^{2}}+{{\left( 4-2 \right)}^{2}}}=\sqrt{{{\left( 4 \right)}^{2}}+{{\left( 2 \right)}^{2}}}\]

\[=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}\]units

Distance between 5(8, 4) and P (2, 1),

\[BP=\sqrt{{{\left( 8-2 \right)}^{2}}{{\left( 4-1 \right)}^{2}}}\]

\[=\sqrt{{{6}^{2}}+{{3}^{2}}}=\sqrt{36+9}\]

\[=\sqrt{45}=3\sqrt{5}\]units

\[\therefore \,\,\,AB=2\sqrt{5}=2AP\]

\[\Rightarrow \,\,\,\,\,AP=\frac{AB}{2}\]

Hence, required condition is \[AP=\frac{AB}{2}\].