## Coordinate Geometry, Straight Line

### Coordinate Geometry, Straight Line

"$P(G,K)$ 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 ?????

### Re: Coordinate Geometry, Straight Line

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

1. Rewrite each equation in its

2. Let the

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

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

6. We have thus established an important condition: the

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 $O$,

There are three other cases with

1. Rewrite each equation in its

**form. Keep the $±$ signs for now.***normal/perpendicular*2. Let the

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

**: $(C_1, C_2)≡(+, +), (+, -), (-, +), (-,-).$***about the signs of the constant pair*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

**, $(na+mb)>0$ for $ψ$ to be obtuse.If $C_1$ and $C_2$ are of***same signs***, $(na+mb)<0$ for $ψ$ to be obtuse.***opposite signs*6. We have thus established an important condition: the

**$$C_1×C_2×(na+mb)>0$$.***obtuse angle will contain the origin if*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 $O$,

**$$f(G, K)×Q(G, K)×(na+mb)>0$$.***will be in the obtuse angle if*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:**$$f(G, K)×Q(G, K)×(na+mb)>0$$**### Re: Coordinate Geometry, Straight Line

Thanks a lot. It was really helpful.