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Important Questions (Part III)
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
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??
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.
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.
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.
I hope it's clear now ^_^
S
Sriman
5 months ago
how u found the locus abantika