A straight line which makes equal intercepts | vedclass

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(B) The equation of a line that makes equal intercepts $a$ on the positive $x$ and $y$ axes is given by $\frac{x}{a} + \frac{y}{a} = 1$,which simplifi

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$\sqrt{2} - 1$

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(B) The equation of a line that makes equal intercepts $a$ on the positive $x$ and $y$ axes is given by $\frac{x}{a} + \frac{y}{a} = 1$,which simplifies to $x + y = a$ (where $a > 0$).The perpendicular distance of this line from the origin $(0, 0)$ is given as $1$ unit.Using the distance formula $\left| \frac{Ax_1 + By_1 + C}{\sqrt{A^2 + B^2}} \right| = d$,we have:$\left| \frac{0 + 0 - a}{\sqrt{1^2 + 1^2}} \right| = 1$$\left| \frac{-a}{\sqrt{2}} \right| = 1$Since $a > 0$,we get $a = \sqrt{2}$.Thus,the equation of the line is $x + y = \sqrt{2}$ ... $(i)$.The second line is given as $y = 2x + 3 + \sqrt{2}$,which can be rewritten as $2x - y = -3 - \sqrt{2}$ ... (ii).To find the intersection point $(x_0, y_0)$,we solve equations $(i)$ and (ii):Adding $(i)$ and (ii):$(x + y) + (2x - y) = \sqrt{2} + (-3 - \sqrt{2})$$3x = -3 \implies x_0 = -1$.Substituting $x_0 = -1$ into $(i)$:$-1 + y_0 = \sqrt{2} \implies y_0 = \sqrt{2} + 1$.Finally,we calculate $2x_0 + y_0$:$2(-1) + (\sqrt{2} + 1) = -2 + \sqrt{2} + 1 = \sqrt{2} - 1$.

$A$ straight line which makes equal intercepts on positive $X$ and $Y$ axes and which is at a distance of $1$ unit from the origin intersects the straight line $y = 2x + 3 + \sqrt{2}$ at $(x_0, y_0)$. Then $2x_0 + y_0$ is equal to

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