IIT JEE Entrance Exam> coordinate geometry...

A line through A(-5,-4) meets the lines x+3y+2=0,2x+y+4=0 and x-y-5=0 at the points B,C ad D respectively. If (15/AB)^2 +(10/AC)^2=(6/AD)^2 .Find the equation of the line.

susmita maiti , 13 Years ago

Grade 11

3 Answers

neeraj agarwal

first you take a line passing through A and having inclination m and then get a b c d and u can solve now.

u cn see also that it will form a right angled triangle .

Last Activity: 13 Years ago

Ankit

Let slope be tanx then using parametric form of line write coordinates of B,C and D in terms of sinx and cosx. Then put points B,C,D in the lines given. You will directly get 15/AB, 10/AC and 6/AD in terms of cosx and sinx. Put in equation given to get tanx = -2/3. Slope known point known so line will be 2x + 3y + 22 = 0

Last Activity: 9 Years ago

saif

A(-5,-4)
x+5/cos@=y+4/sin@=k
x=kcos@-5,y=ksin@-4 hoga n
AB=k1,AC=k2,AD=k3
usi tarah (k1cos@-5,k1sin@-4)lies karega x+3y+2=0 pr
yha se k1 ka value aayega
k1=15/cos@+3sin@ jo isse milega
15/AB=cos@+3sin@ hoga
isi trah 10/AC=2cos@+sin@ aur 6/AD=cos@-sin@ hoga n
(15/AB)^2+(10/AC)^2=(6/AC)^2 m rkhne pr
2+3tan@=0 aayega n
slope hogya -2/3 jo tan@ h
equatn of line y+4=tan@(x+5)
y+4=-2/3(x+5)
2x+3y+22=0 hoga jo line h

Last Activity: 8 Years ago