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 lines passing through the origin $(0,0)$.Since one line passes through $(0,0)$ and $(2,3)$,it

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(B) The equation $ax^2+2hxy+by^2=0$ represents a pair of lines passing through the origin $(0,0)$.Since one line passes through $(0,0)$ and $(2,3)$,its equation is $y - 0 = \frac{3-0}{2-0}(x - 0)$,which simplifies to $3x - 2y = 0$.Since the other line passes through $(0,0)$ and $(4,5)$,its equation is $y - 0 = \frac{5-0}{4-0}(x - 0)$,which simplifies to $5x - 4y = 0$.The joint equation of these two lines is $(3x - 2y)(5x - 4y) = 0$.Expanding this,we get $15x^2 - 12xy - 10xy + 8y^2 = 0$,which is $15x^2 - 22xy + 8y^2 = 0$.Comparing this with $ax^2 + 2hxy + by^2 = 0$,we get $a = 15$,$2h = -22$ (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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