In what ratio does the x axis divides the line segment joining a(5 6) and b(2 -8)

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).

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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).

In what ratio does the x axis divide the join of A2, 3 and B 5, 6? a 2 : 3 b 3 : 5 c 1 : 2 d 2 : 1

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$\frac{6k-3}{k+1}=0$

$6k-3=0$

$k=\frac{3}{6}=\frac{1}{2}$

Thus, x-axis divides the line segment joining the points (2, –3) and (5,6) in the ratio 1:2.

Question 5 Coordinated Geometry - Exercise 7.3

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Answer:

Let the ratio in which x-axis divides the line segment joining (–4, –6) and (–1, 7) = 1: k.

Then,

x-coordinate becomes, \frac{\left(-1-4k\right)}{(k+1)}

y-coordinate becomes, \frac{\left(7-6k\right)}{(k+1)}

Since P lies on x-axis, y coordinate = 0

\frac{\left(7-6k\right)}{(k+1)}=0\\ 7-6k=0\\ k=\frac{7}{6}

Therefore, the point of division divides the line segment in the ratio 6 : 7.

Now, m1 = 6 and m2 = 7

By using the section formula,

x=\frac{\left(m_1x_2+m_2x_2\right)}{(m_1+m_2)}=\frac{\left[6(-1)+7(-4)\right]}{(6+7)}=\frac{\left(-6-28\right)}{13}=-\frac{34}{13}\\ So,\ now\\ y=\frac{\left[6(7)+7(-6)\right]}{(6+7)}=\frac{\left(42-42\right)}{13}=0

Hence, the coordinates of P are (-34/13, 0)

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