Question

Show that the points are A (3, 2,-4),B (9, 8,-10), C(-2,-3, 1) collinear.

Answer 1

To prove three points A (3, 2,-4),B (9, 8,-10), C(-2,-3, 1)  Collinear we will find the equation of line passing through AB and show that point C lies on it.

Equation of line passing through (a,p),(b,q),(c,r) is

$\frac{x-a}{p-a}=\frac{y-b}{q-b}=\frac{z-c}{r-c}$

Equation of line AB is

$\frac{x-3}{9-3}=\frac{y-2}{8-2}=\frac{z+4}{-10+4}$

=$\frac{x-3}{6}=\frac{y-2}{6}=\frac{z+4}{-6}$---(1)

now we will check whether  C(-2,-3, 1) lies on AB or not.

Equation (1) becomes

$\frac{-2-3}{6}=\frac{-3-2}{6}=\frac{1+4}{-6}\\\frac{-5}{6}=\frac{-5}{6}=\frac{5}{-6}$,

Which shows that points A (3, 2,-4),B (9, 8,-10), C(-2,-3, 1)  are Collinear.

 

Answer 2

${ \huge{\boxed{\tt {\color{red}{Answer}}}}}$

‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗

We know that the direction ratios of the line passing through two points P(x1, y1, z1) and Q(x2, y2, z2) are given by:

x2 – x1, y2 – y1, z2 – z1 or x1 – x2, y1 – y2, z1 – z2

Given points are A (2, 3, – 4), B (1, – 2, 3) and C (3, 8, – 11).

Direction ratios of the line joining A and B are:

1 – 2, – 2 – 3, 3 + 4

i.e. – 1, – 5, 7.

The direction ratios of the line joining B and C are:

3 –1, 8 + 2, – 11 – 3

i.e., 2, 10, – 14.

From the above, it is clear that direction ratios of AB and BC are proportional.

That means AB is parallel to BC. But point B is common to both AB and BC.

Hence, A, B, C are collinear points

‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗

Hope it's Helpful.....:)