A straight line passing through a fixed point | vedclass

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(C) Let the coordinates of $A$ be $(p, 0)$ and $B$ be $(0, q)$.The equation of the line in intercept form is $\frac{x}{p} + \frac{y}{q} = 1$.Since the

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$\frac{15}{2}$

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(C) Let the coordinates of $A$ be $(p, 0)$ and $B$ be $(0, q)$.The equation of the line in intercept form is $\frac{x}{p} + \frac{y}{q} = 1$.Since the line passes through $(3, 5)$,we have $\frac{3}{p} + \frac{5}{q} = 1$.For a rectangle formed by $A(p, 0)$,$O(0, 0)$,$B(0, q)$,and $C(x, y)$,the point $C$ must have coordinates $(p, q)$.Thus,$p = x$ and $q = y$.Substituting these into the equation,we get $\frac{3}{x} + \frac{5}{y} = 1$.Multiplying by $xy$,we get $3y + 5x = xy$,or $5x - xy + 3y = 0$.Comparing this with the given form $ax + 2hxy + by = 0$,we identify $a = 5$,$b = 3$,and $2h = -1$,which means $h = -\frac{1}{2}$.Therefore,$a + b + h = 5 + 3 - \frac{1}{2} = 8 - \frac{1}{2} = \frac{15}{2}$.

$A$ straight line passing through a fixed point $(3, 5)$ intersects the coordinate axes at two points $A$ and $B$. If the locus of $C(x, y)$,which forms a rectangle with the points $A, O$ (origin),and $B$,is $ax + 2hxy + by = 0$,then $a + b + h =$

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