Question

Find the ratio in which the line segment joining the points A (3,-3) and B (-2,7) is divided by x axis find the oordinates of the point of division

Answer 1

$\mathbb\blue{The\:Answer\:Is}$


Let the required ratio is k : 1 and the coordinate of point of division is (x , 0)

Given points of line segment are: A(3, -3) and B(-2, 7)

Now, (x , 0) = {(-2*k + 3*1)/(k+1), (7*k + (-3)*1)/(k+1)}

=> (x , 0) = {(-2k + 3)/(k+1), (7k - 3)/(k+1)} ..............1

Now (7k - 3)/(k+1) = 0

=> 7k - 3 = 0

=> k = 3/7

So, the ratio is 3/7 : 1 = 3 : 7


Again from equation 1, we get

     (x , 0) = {(-2 * 3/7 + 3)/(3/7 + 1), 0}

=> (x , 0) = {(-6/7 + 3)/(3/7 + 1), 0}

=> (x , 0) = {(-6 + 3*7)/(3 + 7), 0}

=> (x , 0) = {(-6 + 21)/10, 0}

=> (x , 0) = (15/10, 0}

=> (x , 0) = (3/2, 0}

So, the coordinate of point of division is (3/2 , 0)


$\boxed{Hope\:This\:Helps}$

Answer 2

Heyy mate ❤✌✌❤

Here's your Answer...

⤵️⤵️⤵️⤵️⤵️

Let the ratio be K: 1.

By section Formula,

(-2k + 3) / (k+1) , (7k - 3) / (k+ 1)

since, the point is on X axis.

Therefore, X = 0

7k -3 / k +1 = 0

7k - 3 = 0

7k = 3

k = 3/7
✔✔✔

Therefore, the ratio is k : 1 = 3:7.
✔✔

Putting the value of k = 3/7, we get point of intersection as

{ [ -2(3/7) + 3] ÷ (3/7)+1 , 0 }

=> { [(-6/7) + 3] ÷ (3/7) + 1 , 0 }

=> [(-6+21)/7 ÷ (3+7)/7 , 0 ]

=> [ 15/7 ÷ 10/7 , 0 ]

=> [ 15/10 , 0 ]

=> ( 3/2 , 0 ).✔✔✔