If $m_{1}$ and $m_{2}$ are the roots of the equation $x^{2} + (\sqrt{3} + 2) x + (\sqrt{3} - 1) = 0,$ then the area of the triangle formed by the lines $y = m_{1} x,$ $y = m_{2} x$ and $y = 2,$ is

\sqrt{33} - \sqrt{11}

sq. units

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\sqrt{33} + \sqrt{11}

sq. units

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\sqrt{33} + \sqrt{7}

sq. units

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\sqrt{33} - \sqrt{7}

sq. units

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Solution

The correct option is B $\sqrt{33} + \sqrt{11}$ sq. units
Given lines are $y = m_{1} x, y = m_{2} x$ and $y = 2$

The coordinates of the vertices of the given triangle are
$(0, 0), (\frac{2}{m_{1}}, 2)$ and $(\frac{2}{m_{2}}, 2)$

So, area of the triangle
$= \frac{1}{2} ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & 0 2 / m_{1} & 2 2 / m_{2} & 2 0 & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣$

$= \frac{1}{2} ∣ ∣ ∣ \frac{4}{m_{1}} - \frac{4}{m_{2}} ∣ ∣ ∣ = 2 ∣ ∣ ∣ \frac{m_{2} - m_{1}}{m_{1} m_{2}} ∣ ∣ ∣$

As $m_{1}$ and $m_{2}$ are roots of the equation $x^{2} + (\sqrt{3} + 2) x + (\sqrt{3} - 1) = 0$ , so
$m_{1} + m_{2} = - \sqrt{3} - 2$ and $m_{1} m_{2} = \sqrt{3} - 1$

$| m_{1} - m_{2} | = \sqrt{(m_{1} + m_{2})^{2} - 4 m_{1} m_{2}} \Rightarrow | m_{1} - m_{2} | = \sqrt{(\sqrt{3} + 2)^{2} - 4 (\sqrt{3} - 1)} \Rightarrow | m_{1} - m_{2} | = \sqrt{3 + 4 + 4 \sqrt{3} - 4 \sqrt{3} + 4} \Rightarrow | m_{1} - m_{2} | = \sqrt{11}$

Therefore, area of the triangle
$= 2 ∣ ∣ ∣ \frac{m_{2} - m_{1}}{m_{1} m_{2}} ∣ ∣ ∣ = \frac{2 \times \sqrt{11}}{\sqrt{3} - 1} = \sqrt{11} (\sqrt{3} + 1)$
$= (\sqrt{33} + \sqrt{11})$ sq. units

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Similar questions

Show that the are of the triangle formed by the lines $y = m_{1} x, y = m_{2}$ x and y = c is equal to $\frac{c^{2}}{4} (\sqrt{33} + \sqrt{11})$ , where $m_{1}, m_{2}$ are the roots of the equation $x^{2} + (\sqrt{3} + 2) x + \sqrt{3} - 1 = 0.$

Q. The area bounded by

y = - | x | + 3, x = 0, y = 0

and

x = 2,

Q. Show that the area of the triangle formed by the lines

y = m_{1} x, y = m_{2} x a n d y = c

is equal to

\frac{c^{2}}{4} (\sqrt{33} + \sqrt{11}),

where

m_{1}, m_{2}

are the roots of the equation

x^{2} + (\sqrt{3} + 2) x + \sqrt{3} - 1 = 0

Q. The area of the triangle formed by the lines y = x, x = 6 and y = 0 is

(a) 36 sq. units
(b) 18 sq. units
(c) 9 sq. units
(d) 72 sq. units

Q. Area of the triangle formed by the line

x + y = 3

and angle bisectors of the pair of straight lines

x^{2} - y^{2} + 2 y = 1

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If m1 and m2 are the roots of the equation x2+(√3+2)x+(√3−1)=0, then the area of the triangle formed by the lines y=m1x, y=m2x and y=2, is

If $m_{1}$ and $m_{2}$ are the roots of the equation $x^{2} + (\sqrt{3} + 2) x + (\sqrt{3} - 1) = 0,$ then the area of the triangle formed by the lines $y = m_{1} x,$ $y = m_{2} x$ and $y = 2,$ is