Question

Let\[A \left( {-1, 1} \right)\], \[B\left( {3, 4} \right)\]and \[C\left( {2, 0} \right)\] be given three points. A line\[y = mx\], \[m > 0\], intersects lines \[AC\] and \[BC\] at point \[P\]and \[Q\] respectively. Let\[{A_1}\] and \[{A_2}\] be the areas of \[\Delta ABC\]and \[\Delta PQC\]respectively, such that \[{A_1} = 3{A_2}\], then find the value of \[m\].
A. \[\dfrac{4}{{15}}\]
B. 1
C. 2
D. 3

Answer 1

Hint: In the given question, three vertices of a triangle are given and the line intersects the two sides of the triangle. By using the determinant method, we will find the area of the triangles. Then using a point-slope form of a linear equation, we will find the equations of the sides of a triangle. Solving the equations with the equation of lines, we will get the coordinates of the intersection points. Again, by using the determinant method, we will find the value of \[m\].
Formula Used:
The area of a triangle with vertices \[A \left( {a, b} \right)\], \[B\left( {c, d} \right)\] and \[C\left( {e, f} \right)\] is:
      \[area\left( {\Delta ABC} \right) = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}a&b&1\\c&d&1\\e&f&1\end{array}} \right|\]
A linear equation of a line that is passing through a point \[\left( {{x_1},{y_1}} \right)\] and \[\left( {{x_2},{y_2}} \right)\] :
     \[y - {y_1} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right)\]

Complete step by step solution:
The vertices of the triangle are \[A \left( {-1, 1} \right)\], \[B\left( {3, 4} \right)\]and \[C\left( {2, 0} \right)\].
The equation of a line intersecting the two sides of a triangle is \[y = mx\], \[m > 0\].


Let’s apply the formula of an area of a triangle using determinant.
\[{A_1} = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}{ - 1}&1&1\\3&4&1\\2&0&1\end{array}} \right|\]
\[ \Rightarrow \]\[{A_1} = \dfrac{1}{2}\left| {\left[ { - 1\left( {4 - 0} \right) - 1\left( {3 - 2} \right) + 1\left( {0 - 8} \right)} \right]} \right|\]
\[ \Rightarrow \]\[{A_1} = \dfrac{1}{2}\left| {\left[ { - 4 - 1 - 8} \right]} \right|\]
\[ \Rightarrow \]\[{A_1} = \dfrac{{13}}{2}\]

Apply the formula \[y - {y_1} = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}\left( {x - {x_1}} \right)\] to find the equations of lines \[AC\] and \[BC\].
Equation of line \[AC\] is:
\[y - 0 = \dfrac{{0 - 1}}{{2 - \left( { - 1} \right)}}\left( {x - 2} \right)\]
\[ \Rightarrow \]\[y = \dfrac{{ - 1}}{3}\left( {x - 2} \right)\]
\[ \Rightarrow \]\[x + 3y = 2\]

Equation of line \[BC\] is:
\[y - 0 = \dfrac{{0 - 4}}{{2 - 3}}\left( {x - 2} \right)\]
\[ \Rightarrow \]\[y = 4\left( {x - 2} \right)\]
\[ \Rightarrow \]\[4x - y = 8\]

Let \[\left( {{x_1}, m{x_1}} \right)\]and \[\left( {{x_2}, m{x_2}} \right)\] are the coordinates of the points \[P\]and \[Q\] respectively.
Now, apply the formula of an area of a triangle using determinant.
\[{A_2} = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}{{x_1}}&{m{x_1}}&1\\{{x_2}}&{m{x_2}}&1\\2&0&1\end{array}} \right|\]
\[ \Rightarrow \]\[{A_2} = \dfrac{1}{2}\left| {\left[ {2\left( {m{x_1} - m{x_2}} \right) + 1\left( {m{x_1}{x_2} - m{x_1}{x_2}} \right)} \right]} \right|\]
\[ \Rightarrow \]\[{A_2} = \dfrac{1}{2}\left| {\left[ {2m\left( {{x_1} - {x_2}} \right)} \right]} \right|\]
\[ \Rightarrow \]\[{A_2} = m\left| {{x_1} - {x_2}} \right|\]

Solve equations \[x + 3y = 2\] and \[y = mx\].
\[x + 3mx = 2\]
\[ \Rightarrow \]\[x = \dfrac{2}{{1 + 3m}}\]
Replace \[x\] by \[{x_1}\]
\[ \Rightarrow \]\[{x_1} = \dfrac{2}{{1 + 3m}}\]

Solve equations \[4x - y = 8\] and \[y = mx\].
\[4x - mx = 8\]
\[ \Rightarrow \]\[x = \dfrac{8}{{4 - m}}\]
Replace \[x\] by \[{x_2}\]
\[ \Rightarrow \]\[{x_2} = \dfrac{8}{{4 - m}}\]
We will put the values of \[{x_1}\] and \[{x_2}\] in the equation \[{A_2} = m\left| {{x_1} - {x_2}} \right|\]
\[{A_2} = m\left| {\dfrac{2}{{1 + 3m}} - \dfrac{8}{{4 - m}}} \right|\]
It is given that, \[{A_1} = 3{A_2}\].
\[\therefore \dfrac{{13}}{2} = 3m\left| {\dfrac{2}{{1 + 3m}} - \dfrac{8}{{4 - m}}} \right|\]
\[ \Rightarrow \]\[\dfrac{{13}}{2} = 3m\left| {\dfrac{{8 - 2m - 8 - 24m}}{{\left( {1 + 3m} \right)\left( {4 - m} \right)}}} \right|\]
\[ \Rightarrow \]\[\dfrac{{13}}{2} = 3m\left| {\dfrac{{26m}}{{4 + 11m - 3{m^2}}}} \right|\]
\[ \Rightarrow \]\[\dfrac{1}{{6m}} = \dfrac{{2m}}{{4 + 11m - 3{m^2}}}\]
\[ \Rightarrow \]\[4 + 11m - 3{m^2} = 12{m^2}\]
\[ \Rightarrow \]\[15{m^2} - 11m - 4 = 0\]
\[ \Rightarrow \]\[15{m^2} - 15m + 4m - 4 = 0\]
\[ \Rightarrow \]\[15m\left( {m - 1} \right) + 4\left( {m - 1} \right) = 0\]
\[ \Rightarrow \]\[\left( {15m + 4} \right)\left( {m - 1} \right) = 0\]
\[ \Rightarrow \]\[m = \dfrac{{ - 4}}{{15}}\] or \[m = 1\]
It is given that \[m > 0\].
\[ \Rightarrow \] \[m = 1\]
Hence the correct option is option B
Note: Students are often missed the concept that is \[m > 0\]. Also the area of a triangle will be always positive. Sometimes students take \[{A_1} = - \dfrac{{13}}{2}\] which is wrong. For this reason they do not get correct value of \[m\].