If the lines x+y = 0 , x-y=0 , 2x+3y-6=0 and the sides of the triangle and ( -2 , a ) is an interior point of the triangle then a lies in the range : Answer should be 2 < a < 10/3 . Please put the full solution correctly and without any spam .
Answers
Answer:
2 < a < 10/3
Step-by-step explanation: It is better to have a view on the plotting of these lines (picture added).
Clearly, all the points that can be defined by (-2 , a) can be found in between the lines x + y = 0 and 2x + 3y - 6 = 0.
If we define (-2, a) is general sense, then it can be like x = -2 (purple line). [y or 'a' is variable]
Then, the point where x = -2 meets x + y = 0
⇒ -2 + y = 0
⇒ y = 2
The point where x = -2 meets x + y = 0 is (-2 , 2).
Similarly, the point where x =-2 meets 2x + 3y -6 = 0
⇒ 2(-2) + 3y - 6 = 0
⇒ y = 10/3
The point where x =-2 meet 2x + 3y -6 =0 is (-2, 10/3).
we need all the points that lie on the line x =-2 and in the region between x + y = 0 and 2x + 3y -6 = 0 .
As the end points are (-2, 2) and (-2 , 10/3). [In between x+y=0 and 2x+3y-6 =0]
y lies in between 2 and 10/3
∴ 2 < y < 10/3
y is defined by 'a', so
2 < a < 10/3
Answer:
Step-by-step explanation:
