A variable line passes through a fixed point | vedclass

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(B) Let the equation of the line passing through $(a, b)$ be $y - b = m(x - a)$.For point $A$ (where $y = 0$): $-b = m(x - a) \implies x = a - \frac{b

Vedclass · Accepted Answer

$a/x + b/y = 1$

Vedclass · Answer

(B) Let the equation of the line passing through $(a, b)$ be $y - b = m(x - a)$.For point $A$ (where $y = 0$): $-b = m(x - a) \implies x = a - \frac{b}{m}$. So,$A = (a - \frac{b}{m}, 0)$.For point $B$ (where $x = 0$): $y - b = m(0 - a) \implies y = b - ma$. So,$B = (0, b - ma)$.Let the intersection point of the lines parallel to the axes be $P(h, k)$.Then $h = a - \frac{b}{m}$ and $k = b - ma$.From the first equation,$\frac{b}{m} = a - h \implies m = \frac{b}{a - h}$.Substitute $m$ into the second equation: $k = b - (\frac{b}{a - h})a$.$k = \frac{b(a - h) - ab}{a - h} = \frac{ab - bh - ab}{a - h} = \frac{-bh}{a - h}$.$k(a - h) = -bh \implies ak - kh = -bh \implies ak = kh - bh$.Dividing by $hka$: $\frac{ak}{hka} = \frac{kh}{hka} - \frac{bh}{hka} \implies \frac{1}{h} = \frac{1}{a} - \frac{b}{ka}$.Rearranging gives $\frac{a}{h} + \frac{b}{k} = 1$. Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{a}{x} + \frac{b}{y} = 1$.

$A$ variable line passes through a fixed point $(a, b)$ and meets the coordinate axes at $A$ and $B$. The locus of the point of intersection of lines passing through $A$ and $B$ and parallel to the coordinate axes is:

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