Question

$\left( {5,10} \right),\left( {15,15} \right)$ and $\left( {5,5} \right)$ are the coordinates of vertices A, B and C respectively of $\Delta ABC$ and P is a point on median AD such that AP:PD=2:3, then the ratio of the areas of the triangles PBC and ABC is
A. 2:3
B. 3:4
C. 3:5
D. 4:5

Answer 1

Hint: In the question first we have to find the coordinates of the median and point on the median using midpoint formula i.e., $\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$ and we also use the section formula for finding the coordinates of point on the median which is given by the formula \[\left( {\dfrac{{{m_1}{x_2} + {m_2}{x_1}}}{{{m_1} + {m_2}}},\dfrac{{{m_1}{y_2} + {m_2}{y_1}}}{{{m_1} + {m_2}}}} \right)\], and also using the area of the triangle formula i.e., \[\dfrac{1}{2} \times \] altitude \[ \times \] base, and substitute the given values to get the desired answer.

Complete step by step solution:
Given vertices of triangle ABC are $\left( {5,10} \right),\left( {15,15} \right)$ and $\left( {5,5} \right)$ and P is a point on the median AD such that AP: PD = 2:3, now we have to find the ratio of triangles PBC and ABC.
Using the diagram,

From the diagram, P divides median in the ratio 2:3, and as D is median of the triangle it will be the midpoint of the side BC,
Now using midpoint formulas i.e.,
Coordinates of D $ = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$
As D is the midpoint of BC substitute \[\left( {{x_1},{y_1}} \right) = \left( {15,15} \right)\]and \[\left( {{x_2},{y_2}} \right) = \left( {5,5} \right)\],
\[ \Rightarrow \]Coordinates of D$ = \left( {\dfrac{{15 + 5}}{2},\dfrac{{15 + 5}}{2}} \right)$,
Now simplifying we get,
 \[ \Rightarrow \]Coordinates of D=\[\left( {\dfrac{{20}}{2},\dfrac{{20}}{2}} \right)\],
Now after dividing we get,
\[ \Rightarrow \]Coordinates of D=\[\left( {10,10} \right)\],
Now the diagram will be,

Now, we know that if P divides median in the ratio\[{m_1}:{m_2}\], so now to find the coordinates of P we will make use of the section formula which is given by,
\[\left( {\dfrac{{{m_1}{x_2} + {m_2}{x_1}}}{{{m_1} + {m_2}}},\dfrac{{{m_1}{y_2} + {m_2}{y_1}}}{{{m_1} + {m_2}}}} \right)\],
Here \[{m_1} = 2\], and \[{m_2} = 3\], and the required coordinates are here will be A\[\left( {5,10} \right)\] and D\[\left( {10,10} \right)\], now substituting the values we get,
\[ \Rightarrow \]Coordinates of P\[ = \left( {\dfrac{{2 \times 10 + 3 \times 5}}{{2 + 3}},\dfrac{{2 \times 10 + 10 \times 3}}{{2 + 3}}} \right)\]
Now simplifying we get,
\[ \Rightarrow \]Coordinates of P\[ = \left( {\dfrac{{20 + 15}}{5},\dfrac{{20 + 30}}{5}} \right)\],
\[ \Rightarrow \]Coordinates of P\[ = \left( {\dfrac{{35}}{5},\dfrac{{50}}{5}} \right)\],
Now we get the coordinates of P\[ = \left( {7,10} \right)\],
Now diagram will be,

Now we find the lengths of the medians of both the triangles i.e., AD and PD as we know the coordinates of P and coordinates of D using distance point formula, which is given by,\[\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \].
For AD A\[\left( {{x_1},{y_1}} \right) = \left( {5,10} \right)\], and D\[\left( {{x_2},{y_2}} \right) = \left( {10,10} \right)\], now using the distance-point formula we get,
\[ \Rightarrow \]AD=\[\sqrt {{{\left( {10 - 5} \right)}^2} + {{\left( {10 - 10} \right)}^2}} \]
Now simplifying we get,
\[ \Rightarrow \]AD\[ = \sqrt {{{\left( 5 \right)}^2} + {{\left( 0 \right)}^2}} \],
\[ \Rightarrow \]AD\[ = \sqrt {{{\left( 5 \right)}^2}} \],
Now taking out the square root we get,
\[ \Rightarrow \]AD =5,
Now we have to find the length of PD again by using distance point formula which is given by,
\[\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]
For PD P\[\left( {{x_1},{y_1}} \right) = \left( {7,10} \right)\], and D\[\left( {{x_2},{y_2}} \right) = \left( {10,10} \right)\], now using the distance-point formula we get,
PD=\[\sqrt {{{\left( {10 - 7} \right)}^2} + {{\left( {10 - 10} \right)}^2}} \]
Now simplifying we get,
\[ \Rightarrow \]PD\[ = \sqrt {{{\left( 3 \right)}^2} + {{\left( 0 \right)}^2}} \],
\[ \Rightarrow \]PD\[ = \sqrt {{{\left( 3 \right)}^2}} \],
Now taking out the square root we get,
\[ \Rightarrow \]PD =3,
And now by using the area of the triangle formulas i.e., we know that the area of triangle is given by the formula \[\dfrac{1}{2} \times \]altitude\[ \times \]base,
First for the area of triangle ABC=\[\dfrac{1}{2} \times AD \times BC\]
Area of triangle ABC=\[\dfrac{1}{2} \times 5 \times BC\],
Now area of triangle PBC=\[\dfrac{1}{2} \times PD \times BC\]
\[ \Rightarrow \]Area of triangle PBC=\[\dfrac{1}{2} \times 3 \times BC\],
Now ratio of triangles PBC and ABC is given by,
\[ \Rightarrow \]Ratio = \[ = \dfrac{{\dfrac{1}{2} \times 3 \times BC}}{{\dfrac{1}{2} \times 5 \times BC}}\]
Now eliminating the like terms we get,
\[\therefore \] Ratio of area of triangles PBC and area of triangle ABC = 3:5, Hence, Option (C) is the correct answer.

Note: A median of a triangle is a line segment from a vertex of the triangle to the midpoint of the side opposite that vertex. Because there are three vertices, there are of course three possible medians.