Question

Point a pq and r are collinear on pr and pq: pr =2/3. P is located on the origin q is located at (x,y) and r is located at (-12,0) . What are the values of x and y

Answer 1

The values of x and y are  8 and 0 respectively.

Explanation

As 'p' is located at (0,0) and 'r'  is located at (-12,0) , that means both 'p' and 'r' are on the x-axis.  So point 'q' will be also on the x-axis , and 'q' is located at (x,y)

So, the value of y will be 0. That means the co ordinate of point 'q' is (x, 0)

Now, using the Distance formula, length of 'pq' $= \sqrt{(x-0)^2 +(0-0)^2}= \sqrt{x^2}= x$

and length of  'pr'  $= \sqrt{(-12-0)^2+(0-0)^2}= \sqrt{144}=12$

Given that,   pq : pr =2/3

So....

$\frac{x}{12} =\frac{2}{3}\\ \\ 3x=24\\ \\ x= \frac{24}{3}=8$

Thus, the values of x and y are  8 and 0 respectively.