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Sachin Rana

IVth Year UG, IIT Bombay | YouTuber (143k subs) | Mentored 3 under 100 ranks in JEE Advanced | No. 1 educator for Organic Chemistry

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(13)

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Dev Datt Sharma

a year ago

sir,. u r. marvellous who is providing details for social cause without getting money. this is humanity. thanx

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Elevated Learner

2 months ago

Beautifully designed course, thank you so much.

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Elevated Learner

2 months ago

x + y = a + b

0

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Kk

Kishor kumar jha

3 months ago

x+y=a+b

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Abantika Karmakar

a year ago

The required locus is : x+y = a+b.

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Aditya Singh

a year ago

what is the answer of hw question

Abantika Karmakar

a year ago

I am getting two triangles but I am not sure

Aditya Singh

a year ago

I m getting 4 triangle by solving the equation but when I draw the triangle on axes there is no such triangle are possible

Abantika Karmakar

a year ago

Yes you are right.. as we know it is a right angled triangle therefore we can consider it to be drawn inside a circle with diameter along AB.. therefore two triangles on side ( considering the area) and two on the another side.

Sachin Rana

a year ago

The locus is correct. The homework question has 0 triangles as the answer.

Abantika Karmakar

a year ago

Okay sir.

Aditya Singh

a year ago

thanku abantika

Abantika Karmakar

a year ago

Welcome.. Aditya

Nidhi gupta

a year ago

can you explain in detail abantika
no. of possible triangle ??????

Nidhi gupta

a year ago

and eq of locus too
plz

Nidhi gupta

a year ago

now it got eq of locus but plz tell no. of possible triangle = 0
how??

Abantika Karmakar

a year ago

I'm really sorry.. I don't have any idea how no triangle is possible.. each time I solve I'm getting some value. If I get to Know the correct explanation I'll surely let you know.

Abantika Karmakar

a year ago

Ok yeah.. I got it.. consider the point to be (h,k) then find the perpendicular distance of this point from the line opposite it u will get the height and you Know the base as 5 units.. apply 0.5 * 5 * height and equate with area.

Abantika Karmakar

a year ago

The second equation comes from the condition that sides AC and CB are perpendicular to each other which gives the locus of point C which is a circle. Substitute the value of k from the first equation in this equation and you get the discriminate of the equation to be <0 which gives imaginary value hence no triangle is possible.

Abantika Karmakar

a year ago

I hope it's clear now ^_^

S

Sriman

2 months ago

how u found the locus abantika

Arkya ghosh

a year ago

sir plzz add a video on how to solve questions in which one eq for two lines is given

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