In what ratio is the line segment joining the points 6 3?

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).

Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).

Let P(-1, 6) divides the line segment joining the points A(-3, 10) and B(6, -8) in the ratio K : 1.Then the co-ordinates of P are given by

Here, we have
x1 = -3, y1 = 10
x2 = 6, y2 = -8
and    m1 = K, m2 = 1So, the co-ordinates of P are

But the co-ordinates of P are given as P(-1, 6)

Hence, required ratio = 2 : 7.

Answer

Verified

Hint: To find the ratio in which line joining two points is divided, we can use the section formula. We can substitute all the known coordinates in the formula and then by comparison of corresponding coordinates we can find the required ratio.sectional formula is given as:If the points A and B have coordinates $ \left( {{x_1},{y_1}} \right) $ and $ \left( {{x_2},{y_2}} \right) $ respectively. C (x, y) the point dividing this line in the ratio m:n, then: $ C\left( {x,y} \right) = \left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},{\text{ }}\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right) $

Complete step-by-step answer:

We have been given the coordinates of the two points. Let these points be A and BA (6, 4)B (1, −7)Let the point lying on the x axis where the axis intersects with the line be (x, y). but as this point lies on the x – axis, its y coordinate will be 0. Let this point be XX (x, 0)Let the ratio in which the line gets divided by the axis be k:1

Applying the sectional formula:If the points A and B have coordinates $ \left( {{x_1},{y_1}} \right) $ and $ \left( {{x_2},{y_2}} \right) $ respectively. C (x, y) the point dividing this line in the ratio m:n, then according to the sectional formula: $ C\left( {x,y} \right) = \left( {\dfrac{{m{x_2} + n{x_1}}}{{m + n}},{\text{ }}\dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right) $ Here,$   {x_1} = 6 \\  {x_2} = 1 \\$ $   {y_1} = 4 \\  {y_1} = - 7 \\  $ $   x = x \\  y = 0 \\$ $   m = k \\  n = 1 \\$ Substituting the values, we get: $ X\left( {x,0} \right) = \left( {\dfrac{{k + 6}}{{k + 1}},{\text{ }}\dfrac{{ - 7k + 4}}{{k + 1}}} \right) $ As one of the coordinates on LHS of the equation is 0, so we will use that for comparison.So, comparing the y coordinates of both the sides of the equation, we get: $    \Rightarrow \dfrac{{ - 7k + 4}}{{k + 1}} = 0 \\   \Rightarrow - 7k + 4 = 0 \\   \Rightarrow - 7k = - 4 \\   \Rightarrow k = \dfrac{4}{7} \\   $ The required ratio will be given as: $   k:1 = \dfrac{k}{1} \\   \Rightarrow \dfrac{{\dfrac{4}{7}}}{1}{\text{ }}\left( {k = \dfrac{4}{7}} \right) \\   \Rightarrow 4:7 \\   $ Therefore, the ratio in which the line joining the points (6, 4) and (1, −7) is divided by x-axis is $4:7$ and the correct option is B)

So, the correct answer is “Option B”.

Note: We can also take the ratio is m:n but that would just increase the calculation. When we take the ratio k:1 it means $ \dfrac{{\dfrac{m}{n}}}{1} $ as the ratio of m and n can be equal to any constant k. so taking k:1 as the ratio makes no difference but decrease the calculative portion.

The ratio in terms of fraction can be written as: $ a:b = \dfrac{a}{b} $ If this point which divides the line segment is midpoint, then the ratio in which the line gets divided is 1:1