Show that the area of the triangle formed by the lines y - m1 x- y - m2 x and y - c is equal to c24 -33 - -11- where m1- m2 are the roots of the equation x2 - -3 - 2x - -3 - 1 - 0-

Coordinates of are (x, y) = (o, o) , (x_2, y_2) = (cm_1, c), (x_3, y_3) = (cm_2, c) Area of = 12|det x_1 amp;x_2 amp;x_3 y_1 amp;y ...

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Standard X

Mathematics

Solving a Quadratic Equation by Completion of Squares Method

Show that the...

Question

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

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Solution

coordinates of $△$ are $(x, y) = (o, o), (x_{2}, y_{2}) = (\frac{c}{m_{1}}, c), (x_{3}, y_{3}) = (\frac{c}{m_{2}}, c)$
Area of $△ = \frac{1}{2} ∣ ∣ ∣ ∣ d e t ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & x_{2} & x_{3} y_{1} & y_{2} & y_{3} 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$
$= \frac{1}{2} ∣ ∣ ∣ ∣ ∣ d e t ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & \frac{c}{m_{1}} & \frac{c}{m_{2}} 0 & c & c 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣$
$= \frac{c^{2}}{2} (\frac{1}{m_{1}} - \frac{1}{m_{2}}) = \frac{c^{2}}{2} (\frac{m_{2} - m_{1}}{m_{1} m_{2}})$
$m_{1}, m_{2}$ are the roots of $x^{2} + (\sqrt{3} + 2) x + (\sqrt{3} - 1) = 0$
$∴ m_{1} + m_{2} = \frac{- (\sqrt{3} + 2)}{1}$ $(∵ α + β = - \frac{b}{a})$
$m_{1} m_{2} = \frac{\sqrt{3} - 1}{1}$ $(∵ α β = \frac{c}{a})$
$(m_{2} - m_{1})^{2} = (m_{2} + m_{1})^{2} - 4 m_{1} m_{2}$
$= (- (\sqrt{3} + 2))^{2} - 4 (\sqrt{3} - 1) = 3 + 4 + 4 \sqrt{3} - 4 \sqrt{3} - 4 \sqrt{3} + 4 = 11$
$∴$ Area of $△ = \frac{e^{2}}{2} (\frac{\sqrt{11}}{\sqrt{3} - 1}) = \frac{c^{2}}{2} (\frac{\sqrt{11}}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1})$
$= \frac{c^{2}}{2} (\frac{\sqrt{33} + \sqrt{11}}{2}) = \frac{c^{2}}{4} (\sqrt{33} + \sqrt{11})$

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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. Show that the area of the triangle formed by the lines y = m₁ x, y = m₂ x and y = c is equal to

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

where m₁, m₂ are the roots of the equation

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

Q. 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 :

Q. 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,

Q. A triangle has the lines

y = m_{1} x

and

y = m_{2} x

for two of its sides, where

m_{1}, m_{2}

are the roots of the equation

x^{2} + a x - 1 = 0

, then find the orthocentre of triangle.

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Show that the area of the triangle formed by the lines y=m1 x,y=m2 x and y=c is equal to c24(√33+√11), where m1,m2 are the roots of the equation x2+(√3+2)x+√3−1=0.

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