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The perpendicular from the origin to the line $y=m x+c$ meets it at the point $(-1,2)$. Find the values of $m$ and $c$.
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Verified Answer
Let the perpendicular $O M$ is drawn from the origin to $A B$.
$M$ is the foot of the perpendicular
Slope of $O M=\frac{2-0}{-1-0}=\frac{2}{-1}$;
Slope of $A B=m$
$O M \perp A B \quad \therefore m \times(-2)=-1 \quad \therefore \mathrm{m}=\frac{1}{2}$
$M(-1,2)$ lies on $A B$ whose equation is
$y=m x+c$ or $y=\frac{1}{2} x+c$
$2=\frac{1}{2} \times(-1)+c \Rightarrow c=2+\frac{1}{2}=\frac{5}{2}$
$\therefore \quad m=\frac{1}{2}$ or $c=\frac{5}{2}$
$M$ is the foot of the perpendicular
Slope of $O M=\frac{2-0}{-1-0}=\frac{2}{-1}$;
Slope of $A B=m$
$O M \perp A B \quad \therefore m \times(-2)=-1 \quad \therefore \mathrm{m}=\frac{1}{2}$
$M(-1,2)$ lies on $A B$ whose equation is
$y=m x+c$ or $y=\frac{1}{2} x+c$
$2=\frac{1}{2} \times(-1)+c \Rightarrow c=2+\frac{1}{2}=\frac{5}{2}$
$\therefore \quad m=\frac{1}{2}$ or $c=\frac{5}{2}$
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