Coordinate Geometry, Straight Line

For students of class 11-12 (age 16+)
oro
Posts: 10
Joined: Thu Oct 24, 2013 11:36 pm

Coordinate Geometry, Straight Line

Unread post by oro » Thu Oct 31, 2013 6:41 pm

"$P(G,K)$ বিন্দুটি $f(X,Y)=nX+mY+C_1=0$ এবং $Q(X,Y)=aX+bY+C_2=0$ এর মধ্যবর্তী স্থূল কোণ এর অন্তর্ভুক্ত হবে যদি $f(G,K) \times Q(G,k) \times (na+mb)>0$ হয় " এই condition টার কোনো Proof কি দেয়া যাবে ?????

SMMamun
Posts: 57
Joined: Thu Jan 20, 2011 6:57 pm

Re: Coordinate Geometry, Straight Line

Unread post by SMMamun » Sun Nov 10, 2013 3:01 am

I am not fully solving the problem, but showing you the necessary steps that should be sufficient.

1. Rewrite each equation in its normal/perpendicular form. Keep the $±$ signs for now.

2. Let the normals on the lines from the origin $O(0, 0)$ respectively make angle $α_1$ and $α_2$ with positive $x-$axis. Find cosines and sines of these angles from step 1.

3. If the angle between the normals is $α$, and the angle between the lines in which the origin lies is $ψ$, then $α=α_1˜α_2$ and $α+ψ=180°$.

4. $ψ$ will be obtuse if $α$ is acute, i.e. if $cos(α_1˜α_2)>0$. Expand the cosine.

5. Now consider four possible cases about the signs of the constant pair: $(C_1, C_2)≡(+, +), (+, -), (-, +), (-,-).$
For each of the cases, choose appropriate value of $cosα_1, sinα_1, cosα_2, sinα_2$ from step 2 (i.e. taking either $+$ or $–$ from $±$ based on the sign of corresponding $C_1, C_2$) and put the values in the inequality of step 4.

You would find that when $C_1$ and $C_2$ are of same signs, $(na+mb)>0$ for $ψ$ to be obtuse.If $C_1$ and $C_2$ are of opposite signs, $(na+mb)<0$ for $ψ$ to be obtuse.

6. We have thus established an important condition: the obtuse angle will contain the origin if $$C_1×C_2×(na+mb)>0$$.

7. Now consider a case: "$(C_1, C_2)$ are same signs and origin is in the obtuse angle". So, the point $P(G, K)$ must be in the same angle region as $O(0, 0)$ or in the opposite/vertex angle region to validate the condition. Since $(C_1, C_2)$ are same signs, $f(G, K)$ and $Q(G, K)$ must also be same signs to ensure $P$'s right placement with respect to $O$ and the two lines and consequently in the obtuse angle region. Thus $P$, like $O$, will be in the obtuse angle if $$f(G, K)×Q(G, K)×(na+mb)>0$$.

There are three other cases with opposite signs of $(C_1, C_2)$ and origin being in the acute angle. Try in the same way. The condition will always be the same:
$$f(G, K)×Q(G, K)×(na+mb)>0$$

oro
Posts: 10
Joined: Thu Oct 24, 2013 11:36 pm

Re: Coordinate Geometry, Straight Line

Unread post by oro » Mon Nov 11, 2013 3:06 pm

Thanks a lot. It was really helpful.

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