1) find the equations of the three sides.
2) find the condition for which points are within the triangle.. (see whether it includes or excludes zero)
3) then compare the three lines, u will get the ans i mean the proof...
prove that all pts lying inside triangle formed by pts. (1,3),(5,0),(-1,2) satisfy 3x+2y≥0 and 2x+y-13≥0
1) find the equations of the three sides.
2) find the condition for which points are within the triangle.. (see whether it includes or excludes zero)
3) then compare the three lines, u will get the ans i mean the proof...
arey mujhe abhi poora solve karni hai ??
[2] [2]
dekho......
first find the eqn of all the three sides...
then find out the conditions for which one point will be inside the triangle.
jus do it.. wid pen n paper.. [3]
u will get to know..
see for ex for a point to be inside a circle : x^2 +y^2 < r^2
similarly try to get that condition ..
simply chcek which line consists origin, which do not ... n all...
[aur main zaada nahi bolungi... yeh tumhara doubt thori na tha [3] ... maine isko dhund ke nikali pink karwane ke liye... tumne doubt kyun puchha [7] [4] [3]]
eureka i think the inequality must be 2x+y-13≤0 can u check the question.
is it??
i think its corect... but it was one month ago i did this...
so forgot .... but i think i had got ..
chcek it...
u got y-2x-13=0 line .... but he has asked for y+2x-13=0 line..
Problem is wrong
no point lying inside the triangle will satisfy second inequality
second inequality must be reverse.
if second inequality is reversed then check only for vertices.
if vertices satisfy the inequality then every point inside the triangle must satisfy the inequality...