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 c 2-4-33-11- where m 1- m 2 are the roots ofthe equation x2-3-2 x-3-1-0

Y=m_1 x, y=m_2 x and y=c Vertices of triangle formed by above lines are A(0,0 ), B(c/m_1,c ), C(c/m_2,c ) So Area of triangle when three vertices are given i ...

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Byju's Answer

Standard XII

Mathematics

Area of Triangle Using Coordinates

Show that the...

Question

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

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Solution

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

$Vertices of triangle formed by above lines are$

$A (0, 0), B (\frac{c}{m_{1}}, c), C (\frac{c}{m_{2}}, c)$

$So Area of triangle when three vertices are given is$

$= \frac{1}{2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & 0 & 1 \frac{c}{m_{1}} & 0 & 1 \frac{c}{m_{2}} & 0 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$

$= \frac{1}{2} {x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})}$

$= \frac{1}{2} [∣ ∣ \frac{c^{2}}{m_{1}} - \frac{c^{2}}{m_{2}} ∣ ∣] = \frac{c^{2}}{2} [∣ ∣ \frac{m_{2} - m_{1}}{m_{1} m_{2}} ∣ ∣]$

$G i v e n m_{1} a n d m_{2} are roots of$

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

$Product of roots = m_{1} m_{2} = \sqrt{3} - 1$

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

$= \sqrt{(\sqrt{3} + 2)^{2} - 4 \sqrt{3} + 4} .$

$| m_{2} - m_{1} | = \sqrt{3 + 4 + 4 \sqrt{3} - 4 \sqrt{3} + 4} = \sqrt{11}$

$A r e a = \frac{c^{2}}{2} [\frac{\sqrt{11}}{\sqrt{3} - 1}]$

$Rationalising denominator gives$

$\frac{c^{2}}{4} [\sqrt{33} + \sqrt{11}]$

$Hence proved.$

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Similar questions

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. 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,

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