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(A) Let the foot of the perpendicular from $A(2,4)$ onto the line $x+y-1=0$ be $B(h,k)$.Using the formula for the foot of the perpendicular from $(x_1

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$y=-3x$

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(A) Let the foot of the perpendicular from $A(2,4)$ onto the line $x+y-1=0$ be $B(h,k)$.Using the formula for the foot of the perpendicular from $(x_1, y_1)$ to $ax+by+c=0$:$\frac{h-x_1}{a} = \frac{k-y_1}{b} = -\frac{ax_1+by_1+c}{a^2+b^2}$Substituting the values:$\frac{h-2}{1} = \frac{k-4}{1} = -\frac{2+4-1}{1^2+1^2} = -\frac{5}{2}$From this,we get:$h-2 = -\frac{5}{2} \Rightarrow h = 2 - 2.5 = -0.5 = -\frac{1}{2}$$k-4 = -\frac{5}{2} \Rightarrow k = 4 - 2.5 = 1.5 = \frac{3}{2}$So,the foot of the perpendicular is $B(-\frac{1}{2}, \frac{3}{2})$.The line passes through $(0,0)$ and $(-\frac{1}{2}, \frac{3}{2})$.The slope $m = \frac{\frac{3}{2}-0}{-\frac{1}{2}-0} = \frac{3/2}{-1/2} = -3$.The equation of the line is $y - 0 = -3(x - 0)$,which simplifies to $y = -3x$.

Find the equation of the straight line passing through $(0,0)$ and the foot of the perpendicular from $(2,4)$ onto the line $x+y-1=0$.

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