Question

Show that the following points are collinear :

(i) A (2, -2), B(-3, 8) and C(-1, 4)

(ii) A(-5, 1), B(5,5) and C(10, 7)

(iii) A(5, 1), B(1, -1) and C(11, 4)

(iv) A(8, 1), B(3, -4) and C(2, -5)

Answer 1

To be colinear, the area of the triangle made by these three points should be zero.

i) Let $A(x_1,y_1)=A(2,-2),B(x_2,y_2)=B(-3,8)$ and
$C(x_3,y_3)=C(-1,4)$ be the given points. Now

= $\frac{1}{2}\left(x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right)$

= $\frac{1}{2}\left(2\right(8–4)+(-3\left)\right(4+2)+(-1\left)\right(-2–8\left)\right)$

= $\frac{1}{2}(8-18+10)$

= 0

Hence, the given points are collinear.

ii)

Let $A(x_1,y_1)=A(-5,1)$,
$B(x_2,y_2)=B(5,5)$ and
$C(x_3,y_3)=C(10,7)$ be the given points. Now

=$\frac{1}{2}\left(x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right)$

= $\frac{1}{2}(-5(5–7)+(5\left)\right(7–1)+(10\left)\right(1–5\left)\right)$

= $\frac{1}{2}(10+30–40))$

= 0

Hence, the given points are collinear.

iii)

Let $A(x_1,y_1)=A(5,1)$,
B $(x_2,y_2)=B(1,-1)$ and
C $(x_3,y_3)=C(11,4)$ be the given points. Now

=$\frac{1}{2}\left(x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right)$

= $\frac{1}{2}\left(5\right(-1–4)+(1\left)\right(4–1)+(11\left)\right(1+1\left)\right)$

= $\frac{1}{2}(-25+3+22)$

= 0

Hence, the given points are collinear.

iv)

Let $A(x_1,y_1)=A(8,1)$,
$B(x_2,y_2)=B(3,-4)$ and
$C(x_3,y_3)=C(2,-5)$. be the given points.
Now

= $\frac{1}{2}\left(x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_{-}{y}_2\left)\right)$

= $\frac{1}{2}\left(8\right(-4+5)+(3\left)\right(-5–1)+(2\left)\right(1+4\left)\right)$

= $\frac{1}{2}(8-18+10)$

= 0

Hence, the given points are collinear.