A straight line makes an intercept on the Y-a | vedclass

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(D) Let the $X$-intercept be $a$ and the $Y$-intercept be $2a$.The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{2a} = 1$.Multiply

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$2x + y = \pm \sqrt{5}$

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(D) Let the $X$-intercept be $a$ and the $Y$-intercept be $2a$.The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{2a} = 1$.Multiplying by $2a$,we get $2x + y = 2a$,or $2x + y - 2a = 0$.The perpendicular distance from the origin $(0, 0)$ to the line is given as $1$.The formula for the distance from $(x_1, y_1)$ to $Ax + By + C = 0$ is $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.Substituting the values: $1 = \frac{|2(0) + 1(0) - 2a|}{\sqrt{2^2 + 1^2}}$.$1 = \frac{|-2a|}{\sqrt{5}}$.$|2a| = \sqrt{5}$,which means $2a = \pm \sqrt{5}$.Substituting $2a$ back into the equation $2x + y = 2a$,we get $2x + y = \pm \sqrt{5}$.

$A$ straight line makes an intercept on the $Y$-axis twice as long as that on the $X$-axis and is at a unit distance from the origin. Then the line is represented by the equations:

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