Question

The line segment joining the points P(3,3) and Q(6,-6) is trisected at points A and B such that A is nearer to P .If A lies on the line given by 2x+y+k =0 , find the value of 'k' .( pls solve in copy) and send to me​

Answer 1

Given:

The line segment joining the points P(3,3) and Q(6,-6) is trisected at points A and B such that A is nearer to P . A lies on the line given by 2x+y+k =0.

To find:

The value of k ?

Calculation:

Since PQ line segment is trisected at A and B , such that point A is closer to point P , it means that that PA : AQ = 1 : 2 .

Now, let's find out the coordinate of A :

$\therefore \: x = \dfrac{m(x2) + n(x1)}{m + n}$

$\implies \: x = \dfrac{1(6) + 2(3)}{1 + 2}$

$\implies \: x = \dfrac{12}{3}$

$\implies \: x = 4$

Now, y coordinate :

$\therefore \: y = \dfrac{m(y2) + n(y1)}{m + n}$

$\implies \: y = \dfrac{1( - 6) + 2( 3)}{1 + 2}$

$\implies \: y = \dfrac{ - 6 + 6}{3}$

$\implies \: y = 0$

So, coordinate of A is (4 , 0 ).

Now, since A is on the line 2x + y + k = 0 , we can put the co-ordinates of A in the equation:

$\therefore \: 2x + y + k = 0$

$\implies\: 2(4) + 0 + k = 0$

$\implies\: 8 + k = 0$

$\implies\: k = - 8$

So, value of k = -8.