A straight line makes equal negative intercep | vedclass

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Question

(B) Let the intercepts on the $x$-axis and $y$-axis be $-a$ and $-a$ respectively, where $a > 0$.The intercept form of the line is $\frac{x}{-a} + \fr

Vedclass · Accepted Answer

$x + y + 4\sqrt{2} = 0$

Vedclass · Answer

(B) Let the intercepts on the $x$-axis and $y$-axis be $-a$ and $-a$ respectively, where $a > 0$.The intercept form of the line is $\frac{x}{-a} + \frac{y}{-a} = 1$, which simplifies to $x + y + a = 0$.The perpendicular distance $d$ from the origin $(0, 0)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|C|}{\sqrt{A^2 + B^2}}$.Here, $A = 1, B = 1, C = a$, and $d = 4$.Substituting these values: $4 = \frac{|a|}{\sqrt{1^2 + 1^2}} = \frac{a}{\sqrt{2}}$.Thus, $a = 4\sqrt{2}$.Substituting $a$ back into the equation: $x + y + 4\sqrt{2} = 0$.

$A$ straight line makes equal negative intercepts on the coordinate axes. If the perpendicular distance from the origin to the line is $4 \text{ units}$, then the equation of the line is:

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