Answer
Hint: The section formula for the point (x, y) which divides the line segment joining the points \[({x_1},{y_1})\] and \[({x_2},{y_2})\] in the ratio m:n is given as \[(x,y) = \left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right)\]. Use this formula to find the ratio in which the x-axis divides the line segment joining the points (1, 2) and (2, 3).Complete step-by-step answer:We need to find the ratio in which the x-axis divides the line segment joining the points (1, 2) and (2, 3).Any point on the x-axis has its y coordinate equal to zero. Hence, let us assume a point (x, 0) that divides the line segment joining the points (1, 2) and (2, 3).The section formula for the point (x, y) which divides the line segment joining the points \[({x_1},{y_1})\] and \[({x_2},{y_2})\] in the ratio m:n is given as follows:\[(x,y) = \left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right)\]
Let AB be divided by the x-axis in the ratio k : 1 at the point P.
Then, by section formula the coordination of P are
`p = ((5k+2)/(k+1),(6k-3)/(k+1))`
But P lies on the x-axis; so, its ordinate is 0.
Therefore , `(6k-3)/(k+1) = 0`
`⇒ 6k -3=0 ⇒ 6k =3 ⇒k = 3/6 ⇒ k = 1/2`
Therefore, the required ratio is `1/2:1 `, which is same as 1 : 2
Thus, the x-axis divides the line AB li the ratio 1 : 2 at the point P.
Applying `k=1/2` we get the coordinates of point.
`p((5k+1)/(k+1) , 0)`
`=p((5xx1/2+2)/(1/2+1),0)`
`= p (((5+4)/2)/((5+2)/2),0)`
`= p (9/3,0)`
= p (3,0)
Hence, the point of intersection of AB and the x-axis is P( 3,0).
Page 2
Let AB be divided by the x-axis in the ratio :1 k at the point P.
Then, by section formula the coordination of P are
`p = ((3k-2)/(k+1) , (7k-3)/(k+1))`
But P lies on the y-axis; so, its abscissa is 0.
Therefore , `(3k-2)/(k+1) = 0`
`⇒ 3k-2 = 0 ⇒3k=2 ⇒ k = 2/3 ⇒ k = 2/3 `
Therefore, the required ratio is `2/3:1`which is same as 2 : 3
Thus, the x-axis divides the line AB in the ratio 2 : 3 at the point P.
Applying `k= 2/3,` we get the coordinates of point.
`p (0,(7k-3)/(k+1))`
`= p(0, (7xx2/3-3)/(2/3+1))`
`= p(0, ((14-9)/3)/((2+3)/3))`
`= p (0,5/5)`
= p(0,1)
Hence, the point of intersection of AB and the x-axis is P (0,1).