# If four points $A(6,3),B( - 3,5),C(4, - 2)$ and $D(x,3x)$ are given in such a way , then find the value of x.A.$\dfrac{3}{8}$ or$- \dfrac{{14}}{8}$ B.2 or -3C. $\dfrac{{11}}{8}$ or$- \dfrac{3}{8}$D. None of these

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Hint: First write the formula of the area of a triangle, then substitute the given coordinates in the formula to obtain the area of the triangle DBC and triangle ABC. Then Substitute the obtained areas in the given equation to obtain the value of x.

Formula Used:
Area=$\dfrac{1}{2}\left[ {{x_1}{\rm{\;}}\left( {{y_{2 - }}{\rm{\;}}{y_3}{\rm{\;}}} \right) + {x_2}{\rm{\;}}\left( {{y_3} - {y_1}{\rm{\;}}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]$ , where $({x_1},{y_1}),({x_2},{y_2}),({x_3},{y_3})$ are the vertices of the triangle.

Complete step by step solution:
Substitute $({x_1},{y_1}),({x_2},{y_2}),({x_3},{y_3})$ by $(6,3),( - 3,5),(4, - 2)$ in the formula $\dfrac{1}{2}\left[ {{x_1}{\rm{\;}}\left( {{y_{2 - }}{\rm{\;}}{y_3}{\rm{\;}}} \right) + {x_2}{\rm{\;}}\left( {{y_3} - {y_1}{\rm{\;}}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]$and calculate to obtain the required area.
$\dfrac{1}{2}\left[ {6\left( {5 - ( - 2)} \right) + ( - 3)( - 2 - 3) + 4(3 - 5)} \right]$
$= \dfrac{1}{2}\left[ {42 + 15 - 8} \right]$
$= \dfrac{{49}}{2}$
So, the area of the triangle ABC is $\dfrac{{49}}{2}.$
Substitute $({x_1},{y_1}),({x_2},{y_2}),({x_3},{y_3})$ by $(x,3x),( - 3,5),(4, - 2)$ in the formula $\dfrac{1}{2}\left[ {{x_1}{\rm{\;}}\left( {{y_{2 - }}{\rm{\;}}{y_3}{\rm{\;}}} \right) + {x_2}{\rm{\;}}\left( {{y_3} - {y_1}{\rm{\;}}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]$and calculate to obtain the required area.
$\left| {\dfrac{1}{2}\left[ {x\left( {5 - ( - 2)} \right) + ( - 3)( - 2 - 3x) + 4(3x - 5)} \right]} \right|$ , taking modulus as an area is always positive.
$= \left| {\dfrac{1}{2}\left[ {7x + 6 + 9x + 12x - 20} \right]} \right|$
$= \left| {\dfrac{1}{2}\left[ {28x - 14} \right]} \right|$
$= \left| {14x - 7} \right|$
So, the area of the triangle DBC is $\left| {14x - 7} \right|$.
Substitute $\dfrac{{49}}{2}$for area of the triangle ABC and $\left| {14x - 7} \right|$ for the area of the triangle DBC in the equation to obtain the value of x.
Hence,
$\dfrac{{\left| {14x - 7} \right|}}{{\dfrac{{49}}{2}}} = \dfrac{1}{2}$
$\dfrac{{2(\left| {14x - 7} \right|)}}{{49}} = \dfrac{1}{2}$
$4(\left| {14x - 7} \right|) = 49$
$14x - 7 = \pm \dfrac{{49}}{4}$
Now, take the positive sign,
$14x - 7 = \dfrac{{49}}{4}$
$14x = \dfrac{{49}}{4} + 7$
$x = \dfrac{{77}}{{56}}$
That is, $x = \dfrac{{11}}{8}$ .
Take the negative sign,
$14x - 7 = - \dfrac{{49}}{4}$
$14x = - \dfrac{{49}}{4} + 7$
$x = \dfrac{{ - 21}}{{56}}$
That is, $x = \dfrac{{ - 3}}{8}$

Option ‘C’ is correct

Note: Students often skip the calculation step of showing the areas of the triangle and directly mention the areas doing the calculation in the roughwork, but that is not correct, we need to show the full calculation in the main answer sheet to avoid any type of mistake. They also sometimes forget to take the modulus and deal with only the positive value but as x is unknown so we need to take the modulus sign to calculate the positive as well as the negative value.