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If three points $(h, 0),(a, b)$ and $(0, k)$ lies on a line, show that $\frac{a}{h}+\frac{b}{k}=1$.
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Verified Answer
The given points are $A(h, 0), B(a, b), C(0, k)$, they lie on the same plane.
$\therefore \quad$ Slope of $A B=$ Slope of $B C$
$\therefore \quad$ Slope of $A B=\frac{b-0}{a-h}=\frac{b}{a-h}$;
Slope of $B C=\frac{k-b}{0-a}=\frac{k-b}{-a}$
$\therefore \quad \frac{b}{a-h}=\frac{k-b}{-a}$ or by cross-multiplication
$\begin{array}{ll} & -a b=(a-h)(k-b) \\ \text { or } & -a b=a k-a b-h k+h b \\ \text { or } & 0=a k-h k+h b \\ \text { or } & a k+h b=h k\end{array}$
Dividing by hk $\Rightarrow \frac{a k}{h k}+\frac{h b}{h k}=1$ or $\frac{a}{h}+\frac{b}{k}=1$
Hence proved.
$\therefore \quad$ Slope of $A B=$ Slope of $B C$
$\therefore \quad$ Slope of $A B=\frac{b-0}{a-h}=\frac{b}{a-h}$;
Slope of $B C=\frac{k-b}{0-a}=\frac{k-b}{-a}$
$\therefore \quad \frac{b}{a-h}=\frac{k-b}{-a}$ or by cross-multiplication
$\begin{array}{ll} & -a b=(a-h)(k-b) \\ \text { or } & -a b=a k-a b-h k+h b \\ \text { or } & 0=a k-h k+h b \\ \text { or } & a k+h b=h k\end{array}$
Dividing by hk $\Rightarrow \frac{a k}{h k}+\frac{h b}{h k}=1$ or $\frac{a}{h}+\frac{b}{k}=1$
Hence proved.
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