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```        1.  if a straight line  through the point P(3,4) make an angle 30 degrees with x-axis and meets the line 12x + 5y +10=0 at Q, find the lenghth of PQ.
2. find X if (X,2) is an interior point of triangle ABC formed by
lines A+B=4,3A-7B=8,4A-B=31.```
7 years ago

Askiitians Expert Ankit Jain- IIT Bombay
18 Points
```										Hello Ashwin,
Its a nice question you have posted.
Generic Equation of a straight line :
y- y1 = m(x -x1). For the question above , (x1,y1) = (3,4).  And slope is given to be m =tan 30 = 1/sqrt(3).
So the equation of line becomes :  y-4 = m(x-3) where m is as mentioned above.
Now the intersection point of two lines can be easily found out by just substituting value of y from one equation to another. Use value of sqrt(3) =1.732 and you the following intersection point:
Q= (-1.44,1.44)
From the two points P and Q
the distance PQ can be calculated as =  5.125
Hope this exaplanation solves your query. This question is just computational intensive.
Regards,
Ankit Jain
```
7 years ago
11 Points
```										Actually what you can do is make use of the parametric form which isx-x'/cos theta=y-y'/sin theta=rHere theta=π/6x'=3y'=4Apply the values and find the values of x and yx=6+r√3/2y=r+8/2Using equation 12x+5y+10=0r=132/12√3+5Thus this is the distance.
```
one year ago
Vikas TU
6846 Points
```										In second part plot the three eqn.s on x -y axis and discover by satifying the origin in those three eqns.u will get as: 3A -7B >=8A + B >= 44A – B  satify the point (X,2) in all these three eqns.we get,X>= 22/3 …............(1)X>=2/3. …...........(2)X TAKING the intersection of (1) (2) and (3)we get,X belongs to [22/3, 33/4]
```
one year ago
Kj
19 Points
```										This question can easily be solved by using parametric eqn firstly find value of x and y in the form of r then put the values in the given eq
```
3 months ago
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