betweine if the points (1,5), (2, 3) and (2,-11) are collinear1,5), (2

1.	betweine if the points (1,5), (2, 3) and (2,-11) are collinear1,5), (2, 3) and (-2,-11) are collinear.
Answer» Let A = (1, 5), B = (2, 3) and C = (-2, -11) Thus, distance between AB by distance formula √(x2−x1)2+(y2−y1)2 Here, x1= 1, x2= 2, y1= 5 and y2= 3 So,AB=√(2−1)2+(3−5)2 AB= √5AB =5units Now, distance between BC by distance formula √(x2−x1)2+(y2−y1) Here, x1= 2, x2= -2, y1= 3 and y2= -11 So,BC=√(−2−2)2+(−11−3)2 =253unitsAnd distance between AC by distance formula √(x2−x1)2+(y2−y1)2 Here, x1= 1, x2= -2, y1= 5 and y2= -11 So,AC=√(−2−1)2+(−11−5)2 =265units Now, for collinear lines AB+BC=AC After substituting the values of AB, BC and AC √5+√53≠√265 Thus, given points are non-collinear.

betweine if the points (1,5), (2, 3) and (2,-11) are collinear1,5), (2, 3) and (-2,-11) are collinear.

Answer»

Let A = (1, 5), B = (2, 3) and C = (-2, -11)

Thus, distance between AB by distance formula

√(x2−x1)2+(y2−y1)2

Here, x1= 1, x2= 2, y1= 5 and y2= 3

So,AB=√(2−1)2+(3−5)2

AB= √5AB =5units

Now, distance between BC by distance formula

√(x2−x1)2+(y2−y1)

Here, x1= 2, x2= -2, y1= 3 and y2= -11

So,BC=√(−2−2)2+(−11−3)2

=253unitsAnd distance between AC by distance formula

√(x2−x1)2+(y2−y1)2

Here, x1= 1, x2= -2, y1= 5 and y2= -11

So,AC=√(−2−1)2+(−11−5)2

=265units

Now, for collinear lines

AB+BC=AC

After substituting the values of AB, BC and AC

√5+√53≠√265

Thus, given points are non-collinear.