If one of the lines represented by ax^2+2hxy+ | vedclass

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(B) The equation $ax^2+2hxy+by^2=0$ represents a pair of straight lines passing through the origin $(0,0)$.Line $1$ passes through $(0,0)$ and $(2,3)$

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$1$

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(B) The equation $ax^2+2hxy+by^2=0$ represents a pair of straight lines passing through the origin $(0,0)$.Line $1$ passes through $(0,0)$ and $(2,3)$. Its equation is $y - 0 = \frac{3-0}{2-0}(x - 0)$ $\Rightarrow y = \frac{3}{2}x$ $\Rightarrow 3x - 2y = 0$.Line $2$ passes through $(0,0)$ and $(4,5)$. Its equation is $y - 0 = \frac{5-0}{4-0}(x - 0)$ $\Rightarrow y = \frac{5}{4}x$ $\Rightarrow 5x - 4y = 0$.The combined equation is $(3x - 2y)(5x - 4y) = 0$.Expanding this,we get $15x^2 - 12xy - 10xy + 8y^2 = 0 \Rightarrow 15x^2 - 22xy + 8y^2 = 0$.Comparing with $ax^2 + 2hxy + by^2 = 0$,we have $a = 15$,$2h = -22$ (so $h = -11$),and $b = 8$.Therefore,$a + 2h + b = 15 - 22 + 8 = 1$.

If one of the lines represented by $ax^2+2hxy+by^2=0$ passes through $(2,3)$ and the other passes through $(4,5)$,then $a+2h+b$ equals

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