Q. The combined equation of three sides of a triangle is (x^2 - y^2)(2x + 3y - 6) = 0. If (-2,a) is an interior point and (b,1) is an exterior point of the triangle then
A) 2 < a < 10/3
B)-2 < a < 10/3
C)-1 < b < 9/2
D)-1 < b < 1
Ans: A,D but how??
The combined equation of three sides of the triangle is
(x^2 - y^2)(2x + 3y - 6) = 0.
By factorizaqtion, we get the three equations as:
x+y=0 --- (i)
x-y=0 --- (ii)
and, 2x + 3y – 6 =0 `rArr` 2x+3y=6 --- (iii)
Let the three sides of the triangle be AB, BC and CA, so that The equation of AB is x+y=0 of BC is x-y=0 and that of CA is 2x+3y=6.
Solving the equations of AB and BC, we get the coordinates of point B, which is (0, 0).
Similarly, solving the equations of BC and CA, we get the coordinates of point C, which is (6/5, 6/5),
and solving the equations of CA and AB, we get the coordinates of point A, which is (-6, 6).
You can help yourself with a diagram of a triangle with coordinates of A, B and C approximately close to the values obtained here, with respect to the coordinate axes.
Now considering point P(-2, a), which is interior to the triangle, it is clear from the diagram that sides AB and AC cut the line x=-2 inside the triangle while side CB cuts it outside. For checking interiority, we have to consider sides AB and AC only.
Putting the value of x=-2,
in AB, -2+y=0 `rArr` y=2,
and in AC, 2*(-2)+3y=6 `rArr` y=10/3.
So, condition for interiority of point P(-2, a) is 2<a<10/3, i.e. option A).
Again, considering point Q(b, 1), which is exterior to the triangle,
sides AB and BC cut the line y=1 inside the triangle while side AC cuts it outside.
So in order to check exteriority of Q, we have to consider sides AB and BC only. Putting the value of y=1,
in AB, x+1=0 `rArr` x=-1
and in BC, x-1=0 `rArr` x=1
So, condition at which the point Q(b, 1) lies inside this triangle is -1<b<1.
Beyond this range of b, the point Q is outside the triangle. Only option D satisfies this criteria of exteriority. Therefore, correct option for exteriority of point Q(b, 1) is option D).