Question

Four points A(6, 3), B(–3, 5), C(4, –2) and D(x, 3x) are given such that \[\dfrac{{\Delta DBC}}{{\Delta ABC}} = \dfrac{1}{2}\], then find the value of x.

Answer 1

Hint:- We have to only find the area of the two triangles \[\Delta DBC\] and \[\Delta ABC\] using formula \[\dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]\], where \[\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right),\left( {{x_3},{y_3}} \right)\] are the coordinates of the vertices of the triangle.

Complete step-by-step answer:

As we know, we are given vertices of the triangle \[\Delta DBC\] and \[\Delta ABC\], but the coordinates of point D are in terms of x. So, let us first find the area of the triangle \[\Delta DBC\] and \[\Delta ABC\] using the formula \[Ar\left( {\Delta XYZ} \right) = \dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]\] where \[\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right),\left( {{x_3},{y_3}} \right)\] are the coordinates of the points X, Y and Z respectively.

So, according to the formula the area of triangle \[\Delta DBC\] will be,

\[Ar\left( {\Delta DBC} \right) = \dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]\]

(where \[\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right),\left( {{x_3},{y_3}} \right)\] are the coordinates of vertices D, B and C respectively.)

So, \[Ar\left( {\Delta DBC} \right) = \dfrac{1}{2}\left[ {x\left( {5 - \left( { - 2} \right)} \right) +

 \left( { - 3} \right)\left( {\left( { - 2} \right) - 3x} \right) + 4\left( {3x - 5} \right)} \right] =

\dfrac{1}{2}\left[ {7x + 6 + 9x + 12x - 20} \right]\]

\[Ar\left( {\Delta DBC} \right) = \dfrac{1}{2}\left[ {28x - 14} \right] = 14x - 7\]

Now we had to find the area of triangle \[\Delta ABC\]

\[Ar\left( {\Delta ABC} \right) = \dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left(

{{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]\]

(where \[\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right),\left( {{x_3},{y_3}} \right)\] are

the coordinates of vertices A, B and C respectively.)

So, \[Ar\left( {\Delta ABC} \right) = \dfrac{1}{2}\left[ {6\left( {5 - \left( { - 2} \right)} \right) +

\left( { - 3} \right)\left( {\left( { - 2} \right) - 3} \right) + 4\left( {3 - 5} \right)} \right] =

\dfrac{1}{2}\left[ {42 + 15 - 8} \right] = \dfrac{{49}}{2}\]

Now as it is given in the question that,

\[\dfrac{{\Delta DBC}}{{\Delta ABC}} = \dfrac{1}{2}\] ………………………(1)

So, putting the value of \[\Delta DBC\] and \[\Delta ABC\] from above equation to equation (1). We get,

\[\dfrac{{14x - 7}}{{\dfrac{{49}}{2}}} = \dfrac{{2\left( {14x - 7} \right)}}{{49}} = \dfrac{{28x -

14}}{{49}} = \dfrac{1}{2}\]

Now cross multiplying the above equation. We get,

2(28x – 14) = 49

56x – 28 = 49

Now adding 28 to both the sides of the above equation. We get,

56x = 77

Dividing both sides of the above equation by 56. We get,

\[x = \dfrac{{77}}{{56}} = \dfrac{{11}}{8}\]

Hence, the value of x will be \[\dfrac{{11}}{8}\] such that the \[\dfrac{{\Delta DBC}}{{\Delta

ABC}} = \dfrac{1}{2}\].

Note:- Whenever we come up with this type of problem then first, we should find the area of both the triangles in the numerator and the denominator of the given equation using formula which states that if XYZ is a triangle and \[\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right),\left( {{x_3},{y_3}} \right)\] are the coordinates of vertices X, Y and Z respectively. Then the area of the triangle XYZ will be equal to \[\dfrac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]\]. After that we will put the area of both the triangles in the given equation and then solve the equation by cross-multiplying LHS and RHS to get the required value of x. This will be the easiest and efficient way to find the solution of the problem.