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  • 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 also lies on the line given by 

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 also lies on the line given by 2x-y+k=0, find the value of k.

 

 

 
 
 
 
 

Answers (1)

In the above diagram,

AP=PQ=QB=1.

A=(2,1) and  B=(5,-8).

AP:AB= 1:3 (given)

\frac{AP}{AP+PB}= \frac{1}{3}

3AP=AP+PB

2AP=PB

\frac{AP}{PB}=\frac{2}{1}.....(1)

So P divides AB in ratio 1:2,

Section formula = \left [ \frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2} \right ]

From eqn(1)

x_1=2,y_1=1,x_2=5,y_2=-8,m_1=1,m_2=2

Putting values in section formula

Coordinates of P = \left [ \frac{1(5))+2(2))}{1+2},\frac{1(-8)+2(1)}{1+2} \right ]

=\left [ \frac{5+4}{3},\frac{-8+2}{3} \right ]

=\left [ \frac{9}{3},\frac{-6}{3} \right ]

=\left ( 3,-2 \right )

P(3,-2) lies on the line 2x-y+k=0

Putting value in eqn

2x-y+k=0

2(3)-(-2)+k=0

6+2+k=0

k=-8

Hence the value of k is -8.

 

 

Posted by

Safeer PP

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