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Question

(A) Let the equation of the lines passing through the origin be $y = mx$. Since the perpendicular distance from the point $({x_1}, {y_1})$ to the line

Vedclass · Accepted Answer

$(x{y_1} - y{x_1})^2 = {d^2}({x^2} + {y^2})$

Vedclass · Answer

(A) Let the equation of the lines passing through the origin be $y = mx$. Since the perpendicular distance from the point $({x_1}, {y_1})$ to the line $mx - y = 0$ is $d$,we have: $\frac{|m{x_1} - {y_1}|}{\sqrt{{m^2} + 1}} = d$ Squaring both sides,we get: $(m{x_1} - {y_1})^2 = {d^2}({m^2} + 1)$ Since $m = \frac{y}{x}$,substituting this into the equation: $(\frac{y}{x}{x_1} - {y_1})^2 = {d^2}((\frac{y}{x})^2 + 1)$ Multiplying by ${x^2}$: $(y{x_1} - x{y_1})^2 = {d^2}({y^2} + {x^2})$ Thus,the equation is $(x{y_1} - y{x_1})^2 = {d^2}({x^2} + {y^2})$.

If the distance of two lines passing through the origin from the point $({x_1}, {y_1})$ is $d$,then the equation of the lines is:

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