Find the ratio in which the …

Let P(a, 0) be the point which divides the line segment joining A(1, -3) and B(4,5) in ratio K: 1
Using section formula we get,

{tex}(a,0)=( \frac{K\times4+1\times 1}{K+1}, \frac{K\times5+1\times-3}{K+1}){/tex}

{tex}(a,0)=( \frac{4K+1}{K+1}, \frac{5K-3}{K+1}){/tex}
a = {tex}\frac{4 \mathrm{K}+1}{\mathrm{K}+1}{/tex} and 0 = {tex}\frac{5 \mathrm{K}-3}{\mathrm{K}+1}{/tex}

{tex}\Rightarrow{/tex} {tex}\frac{5K-3}{K+1} =0{/tex} [ taking y-coordinate]
{tex}\Rightarrow{/tex} 5K-3=0
{tex}\Rightarrow{/tex} K = {tex}\frac{3}{5}{/tex}
The required ratio is 3: 5.
Put the value of k in x-coordinate, we get,
{tex}a =\frac{4 \mathrm{(\frac{3}{5})}+1}{\mathrm{(\frac{3}{5})}+1}{/tex}
{tex}a =\frac{\mathrm{\frac{12+5}{5}}}{\mathrm{\frac{3+5}{5}}} =\frac{\frac{17}{5}}{\frac{8}{5}}=\frac{17}{5}\times \frac{5}{8}=\frac{17}{8}{/tex}
Hence, point P is ({tex}\frac{17}{8}{/tex}, 0).