1
$\begingroup$

If the points with vertices $(p_1,q_1)$ , $(p_2,q_2)$ and $(p_1+p_2,q_1+q_2)$ are collinear , show that $p_1q_2=p_2q_1$

  • 0
    There are all two-dimensional points. – 2017-02-20
  • 0
    @amd thanks sir – 2017-02-20

3 Answers 3

1

The line through $(p_1,q_1)$ and $(p_2,q_2)$ is given by

$$(p_1,q_1)+k(p_1-p_2,q_1-q_2),\quad \text{$k$ is a scalar number}$$

once $(p_1+p_2,q_1+q_2)$ is on the line then:

$$(p_1,q_1)+k(p_1-p_2,q_1-q_2)=(p_1+p_2,q_1+q_2)$$

what give us:

$$p_1+k(p_1-p_2)=p_1+p_2 \to k(p_1-p_2)=p_2\\ q_1+k(q_1-q_2)=q_1+q_2 \to k(q_1-q_2)=q_2$$

Isolating $k$ we have:

$$\frac{p_2}{p_1-p_2}=\frac{q_2}{q_1-q_2}\to p_2(q_1-q_2)=q_2(p_1-p_2)\\ p_2q_1-p_2q_2=p_1q_2-p_2q_2 \to p_2q_1=p_1q_2$$

  • 0
    p2/(p1-p2)=q2/(q1-q2) Is what I got finally, which doesn't match my answer. Please help – 2017-02-20
  • 0
    also, could you help me with a more simplified method if possible? – 2017-02-20
  • 0
    @user5183360: please, take another look – 2017-02-20
  • 0
    sorry, my bad, also could you think of a simpler method for me? – 2017-02-20
1

Equating the slope of the lines between $(p_1,q_1), (p_2,q_2)$ and $(p_2,q_2), (p_1+p_2,q_1+q_2)\,$, respectively, gives (assuming $p_2 \ne p_1\ne 0$):

$$ \require{cancel} \frac{q_2-q_1}{p_2-p_1} = \frac{(q_1+\bcancel{q_2})-\bcancel{q_2}}{(p_1+\cancel{p_2})-\cancel{p2}} \;\;\iff\;\; p_1(q_2-\cancel{q_1}) = (p_2-\cancel{p_1})q_1 \;\;\iff\;\; p_2q_1=p_1q_2 $$

The remaining (trivial) cases $p_2=p_1$ or $p_1=0$ need to be treated separately.

  • 1
    thank u so much finally you answered in the simplest way ! – 2017-02-20
0

Let's try:

The equation of a straight line that passes through $(p_1,q_1)$ and $(p_2,q_2)$:

$y - q_1 = \frac { q_2 - q_1}{p_2 - p_1} ( x - p_1)$ .

(Standard formula).

This line also passes through$ (p_1 + p_2, q_1 + q_2)$.

Insert these values into the above equation of the line:

$q_1 + q_2 - q_1 = \frac{q_2 - q_1}{p_2 - p_1} ( p_1 +p_2 - p_1)$;

$q_2 = \frac{q_2 - q_1}{p_2 - p_1}(p_2)$;

$(p_2 - p_1) q_2 = p_2 ( q_2- q_1)$;

$p_1 q_2 = p_2 q_1$ .