The line AB cuts off equal intercepts 2a from | vedclass

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(B) The equation of the line $AB$ with intercepts $2a$ on both axes is given by the intercept form: $\frac{x}{2a} + \frac{y}{2a} = 1$,which simplifies

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$x+y=a$

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(B) The equation of the line $AB$ with intercepts $2a$ on both axes is given by the intercept form: $\frac{x}{2a} + \frac{y}{2a} = 1$,which simplifies to $x + y = 2a$. Let the coordinates of any point $P$ on the line $AB$ be $(2h, 2k)$. Since $P$ lies on the line $x + y = 2a$,we have $2h + 2k = 2a$,which simplifies to $h + k = a$. Perpendiculars $PR$ and $PS$ are drawn to the axes,so $R$ is $(2h, 0)$ and $S$ is $(0, 2k)$. The mid-point of $RS$ is given by $(\frac{2h+0}{2}, \frac{0+2k}{2}) = (h, k)$. Let the coordinates of the mid-point be $(x, y)$,so $x = h$ and $y = k$. Substituting these into the relation $h + k = a$,we get $x + y = a$. Thus,the locus of the mid-point of $RS$ is $x + y = a$.

The line $AB$ cuts off equal intercepts $2a$ from the axes. From any point $P$ on the line $AB$,perpendiculars $PR$ and $PS$ are drawn to the axes. The locus of the mid-point of $RS$ is

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