Find the ratio in which line …

Let A(-2, -3) and B(5, 6) be the given points.

1. Suppose x-axis divides AB in the ratio k:1 at point P
   Then, the coordinates of the point of division are
   {tex}P \left[ \frac { 5 k - 2 } { k + 1 } , \frac { 6 k - 3 } { k + 1 } \right]{/tex}
   Since P lies on x-axis, and y-coordinates of every point on x-axis is zero.
   {tex}\therefore \quad \frac { 6 k - 3 } { k + 1 } = 0{/tex}
   {tex}\Rightarrow{/tex} 6k - 3=0
   {tex}\Rightarrow{/tex} 6k = 3
   {tex}\Rightarrow{/tex} {tex}k = \frac { 3 } { 6 } \Rightarrow k = \frac { 1 } { 2 }{/tex}
   Hence, the required ratio is 1:2.
   Putting {tex}K = \frac { 1 } { 2 }{/tex}in the coordinates of P.
   We find that its coordinates are {tex}\left( \frac { 1 } { 3 } , 0 \right).{/tex}
2. Suppose y-axis divides AB in the ratio k:1 at point Q.
   Then, the coordinates of the point of division are
   {tex}Q \left[ \frac { 5 k - 2 } { k + 1 } , \frac { 6 k - 3 } { k + 1 } \right]{/tex}
   Since, Q lies on y-axis, and x-coordinates of every point on y-axis is zero.
   {tex}\therefore \quad \frac { 5 k - 2 } { k + 1 } = 0{/tex}
   {tex}\Rightarrow{/tex} 5k - 2 = 0​​​​​​​
   {tex}\Rightarrow \quad k = \frac { 2 } { 5 }{/tex}
   Hence, the required ratio is {tex}\frac { 2 } { 5 } : 1 = 2 : 5{/tex}
   Putting {tex}K = \frac { 2 } { 5 }{/tex}in the coordinates of Q.
   We find that the coordinates are {tex}\left( 0 , \frac { - 3 } { 7 } \right){/tex}​​​​​​​