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Question

(A) The given equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$.This can be rewritten as $\frac{bx + ay}{ab} = 1$,which simplifies to $bx + ay -

Vedclass · Accepted Answer

$\frac{ab}{\sqrt{a^2 + b^2}}$

Vedclass · Answer

(A) The given equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$.This can be rewritten as $\frac{bx + ay}{ab} = 1$,which simplifies to $bx + ay - ab = 0$.The formula for the perpendicular distance $d$ from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.Here,the point is the origin $(0, 0)$,so $x_1 = 0$ and $y_1 = 0$.Substituting the values into the formula: $d = \frac{|b(0) + a(0) - ab|}{\sqrt{b^2 + a^2}}$.$d = \frac{|-ab|}{\sqrt{a^2 + b^2}} = \frac{|ab|}{\sqrt{a^2 + b^2}}$.Thus,the length of the perpendicular is $\frac{|ab|}{\sqrt{a^2 + b^2}}$.

Find the length of the perpendicular drawn from the origin to the line $\frac{x}{a} + \frac{y}{b} = 1$.

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