A) \[52x+89y+519=0\]
B) \[52x+59y-519=0\]
C) \[89x+52y+519=0\]
D) \[89x+52y-519=0\]
Correct Answer: A
Solution :
| Slopes of the lines \[AB\equiv 4x+y=1\]and \[BC\equiv 3x-4y+1=0\]are \[-4\] and \[\frac{3}{4}\] respectively | |
| If \[\alpha \] be the angle between AB and BC, then | |
 |
|
| \[\tan \alpha =\frac{-4-\frac{3}{4}}{1-4\left( \frac{3}{4} \right)}=\frac{19}{8}\] | ...(i) |
| Given, \[AB=\text{ }AC\] | |
| \[\Rightarrow \,\,\,\angle ABC=\angle ACB=\alpha \] | |
| Thus the line A C also makes an angle \[\alpha \] with BC. | |
| If m be the slope of the line AC, then its equation is, | |
| \[y+7=m(x-2)\] | ...(ii) |
| Now \[\tan \alpha =\pm \left[ \frac{m-\frac{3}{4}}{1+m.\frac{3}{4}} \right]\Rightarrow \frac{19}{8}=\pm \frac{4m-3}{4+3m}\] | |
| \[\Rightarrow \,\,m=-4\] or \[-\frac{52}{89}\]. | |
| But slope of AB is \[-4,\] therefore slope of AC is \[-\frac{52}{89}\]. | |
| Hence, the equation of line AC given by (ii) is \[52x+89y+519=0\] | |
You need to login to perform this action.
You will be redirected in
3 sec
