 Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
`        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.`
7 years ago

```							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 .
```
7 years ago
```							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
```
3 years ago
```							A(-5,-4)x+5/cos@=y+4/sin@=kx=kcos@-5,y=ksin@-4 hoga nAB=k1,AC=k2,AD=k3usi tarah (k1cos@-5,k1sin@-4)lies karega x+3y+2=0 pryha se k1 ka value aayegak1=15/cos@+3sin@ jo isse milega15/AB=cos@+3sin@ hogaisi 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 nslope hogya -2/3 jo tan@ hequatn of line y+4=tan@(x+5)y+4=-2/3(x+5)2x+3y+22=0 hoga jo line h
```
2 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on IIT JEE Entrance Exam

View all Questions »  ### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution  ### Course Features

• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions