JEE MAIN - Mathematics (2021 - 16th March Evening Shift - No. 15)
Let A($$-$$1, 1), B(3, 4) and C(2, 0) be given three points.
A line y = mx, m > 0, intersects lines AC and BC at point P and Q respectively. Let A1 and A2 be the areas of $$\Delta$$ABC and $$\Delta$$PQC respectively, such that A1 = 3A2, then the value of m is equal to :
A line y = mx, m > 0, intersects lines AC and BC at point P and Q respectively. Let A1 and A2 be the areas of $$\Delta$$ABC and $$\Delta$$PQC respectively, such that A1 = 3A2, then the value of m is equal to :
1
3
2
$${4 \over {15}}$$
Explanation
_16th_March_Evening_Shift_en_15_1.png)
$${A_1} = \Delta ABC = {1 \over 2}\left| {\left| {\matrix{ { - 1} & 1 & 1 \cr 2 & 0 & 1 \cr 3 & 4 & 1 \cr } } \right|} \right|$$
$${A_1} = {{13} \over 2}$$
Equation of line AC is $$y - 1 = - {1 \over 3}(x + 1)$$
Line AC intersect with line y = mx at P,
Solving we get $$P\left( {{2 \over {3m + 1}},{{2m} \over {3m + 1}}} \right)$$
Equation of line BC is y $$-$$ 0 = 4(x $$-$$ 2)
Line BC intersect with line y = mx at Q,
Solving we get $$Q\left( {{{ - 8} \over {m - 4}},{{ - 8m} \over {m - 4}}} \right)$$
A2 = Area of $$\Delta PQC = {1 \over 2}\left| {\left| {\matrix{ 2 & 0 & 1 \cr {{2 \over {3m + 1}}} & {{{2m} \over {3m + 1}}} & 1 \cr {{{ - 8} \over {m - 4}}} & {{{ - 8m} \over {m - 4}}} & 1 \cr } } \right|} \right|\, = {{{A_1}} \over 3} = {{13} \over 6}$$
$$ \Rightarrow $$ $$ {1 \over 2}\left( {2\left( {{{2m} \over {3m + 1}} + {{8m} \over {m - 4}}} \right) - 1\left( {{{ - 16m} \over {(3m + 1)(m - 4)}} + {{16m} \over {(3m + 1)(m - 4)}}} \right)} \right)$$$$ = \pm {{13} \over 6}$$
$$ \Rightarrow $$ $${{26{m^2}} \over {3{m^2} - 11m - 4}} = \pm {{13} \over 6}$$
$$ \Rightarrow 12{m^2} = \pm (3{m^2} - 11m - 4)$$
taking +ve sign
9m2 + 11m + 4 = 0 (Rejected $$ \because $$ m is imaginary)
taking $$-$$ve sign
15m2 $$-$$ 11m $$-$$ 4 = 0
$$m = 1, - {4 \over {15}}$$
$$ \therefore $$ m = 1 (As given m > 0)
Comments (0)
