If $a + b = 2 h$ then the area of the triangle formed by the lines $a x^{2} + 2 h x y + b y^{2} = 0$ and $x - y + 2$ is $\frac{α (b - a)}{b + a} u n i t$ then $α$ =

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$\begin{matrix} a x^{2} + 2 h x y + b y^{2} = 0 a x^{2} + (a + b) x y + b y^{2} = 0 a x^{2} + a x y + b x y + b y^{2} = 0 a x (x + y) + b y (x + y) = 0 (a x + b y) (x + y) = 0 l_{1} : x + y = 0, a x + b y = 0 : l_{2} a n d l_{3} : x - y + 2 = 0 S o, m_{1} = - 1, m_{3} = 1 A : (- 1, 1) B : (\frac{- 2 a b}{a + b}, \frac{2 a}{a + b}) C : (0, 0) S o, a r e a o f Δ A B C = \frac{1}{2} ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & 0 & 1 - 1 & 1 & 1 - \frac{2 a b}{a + b} & \frac{2 a}{a + b} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = \frac{1}{2} (2 (\frac{b - 1}{b + a})) = \frac{b - a}{b + a} u n i t . \end{matrix}$

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If a+b=2h then the area of the triangle formed by the lines ax2+2hxy+by2=0 and x−y+2 is α(b−a)b+aunit then α =

ax2+2hxy+by2=0ax2+(a+b)xy+by2=0ax2+axy+bxy+by2=0ax(x+y)+by(x+y)=0(ax+by)(x+y)=0l1:x+y=0,ax+by=0:l2andl3:x−y+2=0So,m1=−1,m3=1A:(−1,1)B:(−2aba+b,2aa+b)C:(0,0)So,areaofΔABC=12∣∣ ∣ ∣∣001−111−2aba+b2aa+b1∣∣ ∣ ∣∣=12(2(b−1b+a))=b−ab+aunit.

If $a + b = 2 h$ then the area of the triangle formed by the lines $a x^{2} + 2 h x y + b y^{2} = 0$ and $x - y + 2$ is $\frac{α (b - a)}{b + a} u n i t$ then $α$ =