Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# $\left( {5,10} \right),\left( {15,15} \right)$ 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. 2:3B. 3:4C. 3:5D. 4:5

Last updated date: 17th Jul 2024
Total views: 381.9k
Views today: 10.81k
Verified
381.9k+ views
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.