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Important Questions (Part III)
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This is the last part of the chapter Straight Lines.

Sachin Rana is teaching live on Unacademy Plus

Sachin Rana
IIIrd Year UG, IIT Bombay | YouTuber with over 127k subscribers at Sachin Rana [IITB] | Organic Chemistry & Maths Expert.

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(10)
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x+y=a+b is the answer!
sir plzz add a video on how to solve questions in which one eq for two lines is given
The required locus is : x+y = a+b.
Aditya Singh
a year ago
what is the answer of hw question
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
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.
Okay sir.
Aditya Singh
a year ago
thanku abantika
Welcome.. Aditya
Nidhi gupta
10 months ago
can you explain in detail abantika no. of possible triangle ??????
Nidhi gupta
10 months ago
and eq of locus too plz
Nidhi gupta
10 months ago
now it got eq of locus but plz tell no. of possible triangle = 0 how??
Abantika Karmakar
10 months 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
10 months 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
10 months 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
10 months ago
I hope it's clear now ^_^
Rest solution of Q.1: Putting the values of a' and b' from equation 1 and 2 in equation 3, the locus comes out to be x+y=a+b.
bhaiya.... in second question how can the sum of distances be 0 , even though modulus also comes with distance formula...... I'm stucked in both of the questions plz help
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