A straight line which makes equal intercepts | vedclass

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

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$\sqrt{2}-1$

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(B) The equation of a line which makes equal intercepts $a$ on the positive $x$ and $y$ axes is given by $\frac{x}{a} + \frac{y}{a} = 1$,which simplifies to $x + y = a$. Since the distance of this line from the origin $(0, 0)$ is $1$ unit,we have: $\left| \frac{0 + 0 - a}{\sqrt{1^2 + 1^2}} \right| = 1$ $\left| \frac{-a}{\sqrt{2}} \right| = 1 \implies a = \sqrt{2}$ (since intercepts are on positive axes). So,the equation of the line is $x + y = \sqrt{2}$. We are given another line $y = 2x + 3 + \sqrt{2}$,which can be written as $2x - y = -3 - \sqrt{2}$. To find the intersection point $(x_0, y_0)$,we solve the system: $x + y = \sqrt{2}$ $2x - y = -3 - \sqrt{2}$ Adding the two equations: $3x = -3 \implies x_0 = -1$. Substituting $x_0 = -1$ into $x + y = \sqrt{2}$: $-1 + y_0 = \sqrt{2} \implies y_0 = \sqrt{2} + 1$. Now,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 $1$ 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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