$X(3,5)$, $Y(x,y)$ and $Z(9,-3)$ are collinear points and $XY=5$ units then find $Y$

Question

$X(3,5)$, $Y(x,y)$ and $Z(9,-3)$ are collinear points and $XY=5$ units then find $Y$.

What's the use of the distance $XY$ ?

Answer 1

Your two equations should be:

Equal slopes:

$\frac{y-5}{x-3} = \frac{-3-5}{9-3}$

and

Distance criterion:

$\sqrt{(y-5)^2 +(x-3)^2} = 5$

Solve those simultaneously.

Answering your question more precisely, there are an infinite number of points $Y$ collinear to $X$ and $Z$. Basically, any point on the extended line $XZ$ meets that criterion. The distance criterion "fixes" the position of $Y$ in relation to $X$ so it is essential for a solution. Geometrically, you can imagine the distance criterion as specifying $Y$ to lie on a circle radius $5$ centered on $X$. Where that circle meets the line (may be up to two points) will give the position(s) of $Y$. That's why you need both criteria and two equations.

Answer 2

The equation of line passing through $X$ and $Z$ is found to be $$y=-(4/3 )x +9$$

The distance between $x$ and $y$ is $$\sqrt {(x-3)^2+(y-5)^2} =5$$

Substitute for $y$ and solve for $x$ you get $x=6$ or $x=0$

Thus we have $(x,y)= (6,1)$ or $(0,9)$

Answer 3

Alternate solution:

You can write the equation of the line in parametric form as $(3+6t, 5-8t)$. $t$ ranges between $0$ and $1$ since $t=0$ gives $(3,5)$, and $t=1$ gives $(9,-3)$.

Then you need the distance from $X$ to be $5$. So you get:

$$(3+6t-3)^2 + (5-8t-5)^2 = 25$$

Solve for $t$ and substitute into the equation of the line to find $Y$.