Question

The line segment joining the points A(2,1) And B(5,-8) is trisected at the points P and Q such that P is nearer to A .If P lies on the line 2x-y+k =0,find k .

Answer 1

Solution :

We have provided with a line segment AB where P and Q are it's trisection points . The points are A(2,1) and B(5,-8) . Also from the figure , AP = PQ = QB = 1 (Say) then AP = PB will be 1:2 (m1 :m2) and the points are in the form of A(x1 , y1) and B(x2 , y2)  {Refer to the attachment }

Applying the Section Formula to find the coordinate of PQ  :

⇒ $(\frac{m_1x_2 +m_2x_1}{m_1 +m_2} , \frac{m_1y_2 +m_2y_1}{m_1 + m_2} )$

⇒ $(\frac{(1 *5 + 2*2)}{3} , \frac{(1 * -8 + 2 * 1)}{3} )$

⇒ $(\frac{ 5 + 4}{3} , \frac{-6}{3} )$

⇒ $( \frac{9}{3} , \frac{-6}{3} )$

⇒ $( 3 , -2)$

∴ ( x , y)

According to the question P lies on the line 2x - y + k = 0

⇒ 2 (3) - (-2) + k = 0

⇒ 6 + 2 + k = 0

⇒ k = 2 + 6

⇒ k = -8

Answer 2

Answer:

We have provided with a line segment AB where P and Q are it's trisection points . The points are A(2,1) and B(5,-8) . Also from the figure , AP = PQ = QB = 1 (Say) then AP = PB will be 1:2 (m1 :m2) and the points are in the form of A(x1 , y1) and B(x2 , y2) {Refer to the attachment }