The points on the x-axis whose perpendicular | vedclass

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Question

(A) Let the point on the $x$-axis be $P(h, 0)$.The equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$,which can be written as $bx + ay - ab = 0$.

Vedclass · Accepted Answer

$\left( \frac{a}{b}(b \pm \sqrt{a^2 + b^2}), 0 \right)$

Vedclass · Answer

(A) Let the point on the $x$-axis be $P(h, 0)$.The equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$,which can be written as $bx + ay - ab = 0$.The perpendicular distance from $P(h, 0)$ to the line is given as $a$.Using the distance formula $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$,we have:$a = \frac{|bh + a(0) - ab|}{\sqrt{b^2 + a^2}}$$a\sqrt{a^2 + b^2} = |bh - ab|$$bh - ab = \pm a\sqrt{a^2 + b^2}$$bh = ab \pm a\sqrt{a^2 + b^2}$$h = \frac{a}{b}(b \pm \sqrt{a^2 + b^2})$Thus,the points are $\left( \frac{a}{b}(b \pm \sqrt{a^2 + b^2}), 0 \right)$.

The points on the $x$-axis whose perpendicular distance from the line $\frac{x}{a} + \frac{y}{b} = 1$ is $a$ are

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