What is the point B which lies on the line AC and is 5 units away from A (3, 7) and 2 units away from C (1, 2)?  

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We need to find the point B which lies on AC and which is 5 units away from (3, 7) and 2 units away from (1, 2).

We use the relation that the coordinates of a point on a line between A(x1, y1) and B(x2, y2) and the distance of which from A and B is in the ratio m:n resp. is given by (n*x1 + m*x2)/(m+n) , (n*y1 + m*y2)/(m+n)

The ratio of the distance of the point B from A and C is 5:2. 5+2 is equal to 7.

The coordinates of this point B are given by [(5*1 + 2*3)/7, (5*2 + 2*7)/7]

=> [11 / 7, 24/7].

The required point is (11/7, 24/7)

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

Let (x,y) be the point B.

The Distance  AC =  sqrt {(xA-xB)^2+(yA-yB)^2)^2} = sqrt{(3-1)^2+(7-2)^2} = sqrt{2^2+5^2} = sqrt{4+25} = sqrt29.

By data  B is on AC. AB = 5 , AB = 5 and BC = 2 units.

Therefore by data AB+BC = 5+2 = 7.

Btt the distance from A(3,7) to C(1,2) = sqrt29.

This implies AC =  7  and AC = sqrt29.  So 7 = sqrt 29 which is not possible. Therefore the  given data is inconsistent . Therefore there can not be a point B such that B is on AC abd AB = 5 and BC = 2.

At maximum we can divide the  line AC in the ratio 5 : 2 at a point B. The coordinates of any point B that divides the line joining AC in the ratio 5 : 2 is given by :

(xB, yB) = {2*xA+5*bX)/(2+5), (5*yA+2*yB)/(5+2)}

(xB,yB) = {(2*3+5*1/(5+2)) , (2*7+5*2)/(2+5))}

(xB , yB) = {(11/7 , 24/7).

Therefore the line AC is divide by the point B ( 11/7 ,  24/7) in the ratio 5:2.

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