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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.

6 years ago


Answers : (3)


Hello Ashwin,

Its a nice question you have posted.

Answer to Q no 1.

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.


Ankit Jain

6 years ago
Actually what you can do is make use of the parametric form which is
x-x'/cos theta=y-y'/sin theta=r
Here theta=π/6
Apply the values and find the values of x and y
Using equation 12x+5y+10=0
Thus this is the distance.
6 months ago
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 >=8
A + B >= 4
4A – B
satify the point (X,2) in all these three eqns.
we get,

X>= 22/3 …............(1)
X>=2/3. …...........(2)
TAKING the intersection of (1) (2) and (3)
we get,
X belongs to [22/3, 33/4]
6 months ago

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