Question

the locus of point which is collinear with the points (1,2) and (-2,1) is???
please give solution....

Answer 1

by distance formula =√ ( x\2 - x1) +( y2- y1)
let be 1=x1 , 2= y1, and -2=X2 ,1= y2

√(-2-1)+(1-2) by taking square
√(-3)-1. take square
9-1
ans 8

Answer 2

Answer:

x - 3y + 5 = 0 is the required locus of the point which is collinear with the points (1,2) and (-2,1)

Step-by-step explanation:

Let p(x, y) be any arbitrary point

Given that the points A(1, 2) and B(-2, 1) are collinear

and also the locus of the point p(x, y) is collinear with the above two points

If the points A, B, P are collinear, the slope of the line AB is equal to the slope of the line BP

The Slope of AB = Slope of BP

$\frac{1-2}{-2-1} =\frac{y-1}{x+2}$

- x - 2 = -3y + 3

x - 3y + 5 = 0

which is the required locus of the point which is collinear to two points A(1, 2) and B(-2, 1)