Question

prove that the points p(1,2,3),q(-1,-1,-1) and r(3,5,7)are collinear.​

Answer 1

$\huge\bold{\gray{\sf{Answer:}}}$

$\bold{Explanation:}$

Given:

Three points namely P(1,2,3) Q(-1,-1,-1) and R(3,5,7)

To Prove :

That the points are collinear

P(1,2,3) Q(-1,-1,-1) and R(3,5,7)

Here,we use distance formula between these three given Points.

To be collinear the distance of any two points be equal to the third one.

$x_{1}=1,x_{2}=-1\: \&\: x_{3}=3$

$y_{1}=2,y_{2}=-1\:\&\: y_{3}=5$

$z_{1}=3,z_{2}=-1 \:\&\: z_{3}=7$

We find first PQ

$\sf PQ= \sqrt{ {(x_{2} - x_{1})}^{2} + {(y_{2 }- y_{1})}^{2} + {(z_{2} - z_{1})}^{2} }$

$\sf= > PQ = \sqrt{ {( - 1 - 1)}^{2} + {( - 1 - 2)}^{2} + {( - 1 - 3)}^{2} }$

$\sf= > {( - 2)}^{2} + {( - 3)}^{2} + {( - 4)}^{2}$

$\sf = > 4 + 9 + 16 = \sqrt{29}$

$\sf QR= \sqrt{ {(x_{3} - x_{2})}^{2} + {(y_{3 }- y_{2})}^{2} + {(z_{3} - z_{2})}^{2} }$

$\sf= > QR= \sqrt{ {(3 + 1)}^{2} + {(5 + 1)}^{2} + {(1 + 1)}^{2} }$

$\sf= \sqrt{ {4}^{2} + {6}^{2} + {8}^{2} } = \sqrt{116} = 2 \sqrt{29}$

$\sf PR= \sqrt{ {(x_{3} - x_{1})}^{2} + {(y_{3 }- y_{1})}^{2} + {(z_{3} - z_{1})}^{2} }$

$\sf= > PR = \sqrt{ {(3 - 1)}^{2} + {(5 - 2)}^{2} + {(7 - 3)}^{2} }$

$\sf= > \sqrt{ {(2)}^{2} + {(3)}^{2} + {(4)}^{2} } = \sqrt{4 + 9 + 16} = \sqrt{13 + 16} = \sqrt{29}$

Here ,we see that

$\bold{PQ = \sqrt{29} }$

$\bold{QR=2 \sqrt{29}}$

$\bold{ PR=\sqrt{29}}$

Hence ,it is proved that

$\bold{\red{PQ + PR = QR}}$

$\sf= > \sqrt{29} + \sqrt{29} = 2 \sqrt{29}$

Hence,the above Points are collinear.