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If the intercept of a line between the coordinate axes is divided by the point $(-5,4)$ in the ratio $1: 2$, then find the equation of the line.
Solution:
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Verified Answer
Let $\mathrm{A}(a, 0)$ and $\mathrm{B}(0, b)$ be the point which the line cuts on $\mathrm{X}$ and $\mathrm{Y}$ ax is respectively.
$\therefore$ according to the questions:
$$
\begin{aligned}
&(-5,4)=\left(\frac{2 a+1(0)}{2+1}, \frac{2(0)+1(b)}{2+1}\right) \\
&(-5,4) \equiv\left(\frac{2 a}{3}, \frac{b}{3}\right) \\
&\Rightarrow-5=\frac{2 a}{3}, 4=\frac{b}{3} \quad \Rightarrow a=-\frac{15}{2}, b=12
\end{aligned}
$$
$\therefore$ equation of required line is
$$
\begin{aligned}
&\frac{x}{\left(-\frac{15}{2}\right)}+\frac{y}{(12)}=1 \Rightarrow \frac{2 x}{-15}+\frac{y}{12}=1 \\
&\Rightarrow-8 x+5 y=60 \Rightarrow 8 x-5 y+60=0
\end{aligned}
$$
$\therefore$ according to the questions:
$$
\begin{aligned}
&(-5,4)=\left(\frac{2 a+1(0)}{2+1}, \frac{2(0)+1(b)}{2+1}\right) \\
&(-5,4) \equiv\left(\frac{2 a}{3}, \frac{b}{3}\right) \\
&\Rightarrow-5=\frac{2 a}{3}, 4=\frac{b}{3} \quad \Rightarrow a=-\frac{15}{2}, b=12
\end{aligned}
$$
$\therefore$ equation of required line is
$$
\begin{aligned}
&\frac{x}{\left(-\frac{15}{2}\right)}+\frac{y}{(12)}=1 \Rightarrow \frac{2 x}{-15}+\frac{y}{12}=1 \\
&\Rightarrow-8 x+5 y=60 \Rightarrow 8 x-5 y+60=0
\end{aligned}
$$
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