show that area of a triangle formed by lines y=m1x , y=m2x and y=c is equal to c2/4 (√33 +√11) - Maths - Straight Lines

Question

show that area of a triangle formed by lines y=m1x ,
y=m2x and y=c is equal to c2/4
(√33 + √11) where m1 and
m2 are the roots of the equation x2 +(
√ 3 +2 )x + √
3 - 1 =0

Ankush Jain · Accepted Answer

Given, m1 and m2 are the roots of the equation x2 + (âˆš3 + 2)x + (âˆš 3 - 1) = 0 Therefore, value of m1 and m2 is find out by quadratic formula as:

Area of triangle formed = area of shaded region $= \frac{1}{2} \times b a s e \times h e i g h t = \frac{1}{2} \times [\frac{c}{| m_{1} |} - \frac{c}{| m_{2} |}] \times c = \frac{1}{2} c^{2} [\frac{1}{| m_{1} |} - \frac{1}{| m_{2} |}] = \frac{1}{2} c^{2} [\frac{1}{\frac{\sqrt{3} + 2 - \sqrt{11}}{2}} - \frac{1}{\frac{\sqrt{3} + 2 + \sqrt{11}}{2}}] = \frac{1}{2} c^{2} [\frac{2}{\sqrt{3} + 2 - \sqrt{11}} - \frac{2}{\sqrt{3} + 2 + \sqrt{11}}] = c^{2} [\frac{\sqrt{3} + 2 + \sqrt{11} - (\sqrt{3} + 2 - \sqrt{11})}{(\sqrt{3} + 2)^{2} - 11}] = c^{2} [\frac{2 \sqrt{11}}{3 + 4 + 4 \sqrt{3} - 11}] = c^{2} [\frac{2 \sqrt{11}}{4 \sqrt{3} - 4}] = \frac{c^{2}}{2} [\frac{\sqrt{11}}{\sqrt{3} - 1}] = \frac{11 c^{2}}{2 \times \sqrt{11} \times (\sqrt{3} - 1)} = \frac{11 c^{2}}{2 \times (\sqrt{33} - \sqrt{11})} = \frac{11 c^{2}}{2 \times (\sqrt{33} - \sqrt{11})} \times \frac{\sqrt{33} + \sqrt{11}}{\sqrt{33} + \sqrt{11}} = \frac{11 c^{2} (\sqrt{33} + \sqrt{11})}{2 \times (33 - 11)} = \frac{c^{2} (\sqrt{33} + \sqrt{11})}{4}$

show that area of a triangle formed by lines y=m1x , y=m2x and y=c is equal to c2/4 (√33 + √11) where m1 and m2 are the roots of the equation x2 +( √ 3 +2 )x + √ 3 - 1 =0

show that area of a triangle formed by lines y=m₁x , y=m₂x and y=c is equal to c²/4 (√33 + √11) where m₁ and m₂ are the roots of the equation x² +( √ 3 +2 )x + √ 3 - 1 =0