Question

Show that three points A(1,-2); 3(3,4) and (4,7) lie in a straight line.​

Answer 1

Step-by-step explanation:

1

=slopeofAB=

3−1

4−(−2)

=3

m

2

=slopeofBC=

4−3

7−4

=3

∴m

1

=m

2

∴ AB is parallel to BC and B is common to both AB and BC. Hence, the points. A

(1,−2),B(3,4)andC(4,7) are collinear.

i.e., A, B, C lie on a straight line

Answer 2

Answer:

${x_1 (y_2 – y_3)}$ + ${x_2 (y_3 – y_1)}$ + ${x_3 (y_1 – y_2)}$ = 0

=> 1(4 – 7) + 3(7 + 2) + 4(–2 – 4) = 0

=> – 3 + 27 – 24 = 0

=> – 27 – 27 = 0

=> 0 = 0

L.H.S = R.H.S

Hence, proved.