Intersection between rectangle and line from center point

Question

Consider the following diagram

Given that point A is the origin and the center point for the rectangle, and the coordinates for point B, how do you find the intersection between Line AB and the rectangle. Thanks

Answer 1

Let $B=(x_B,y_B)$, and let $h$ be the half the vertical height of the rectangle (i.e. the y-coordinate of the top side of the rectangle). The coordinates of the intersection $P$ are then simply the coordinates of B scaled by a factor $\frac{h}{y_B}$, i.e. $P=(\frac{h}{y_B}x_B,h)$.

Answer 2

This also depends on the dimensions of the rectangle. Say the rectangle had length $a$ and height $b$ and say that point $B$ had coordinates $(p,q)$. Then the equation of the line joining $A(0,0)$ and $B(p,q)$ is

$y=\frac{q}{p}x$

The equations of the sides of the rectangle are $x=-\frac{b}{2},x=\frac{b}{2},y=-\frac{a}{2},y=\frac{a}{2}$

The intersection between $AB$ and the rectangle is the intersection between $AB$ and any one of the sides of the rectangle. So there are four simultaneous equations to solve

$y=\frac{q}{p}x$ if$-a\le(q)\le(a)$ and $p>\frac{b}{2}$

$x=\frac{b}{2}$

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$y=\frac{q}{p}x$ if$-a\le(q)\le(a)$ and $p<-\frac{b}{2}$

$x=-\frac{b}{2}$

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$y=\frac{q}{p}x$ if$-b\le(p)\le(b)$ and $q>\frac{a}{2}$

$y=\frac{a}{2}$

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$y=\frac{q}{p}x$ if$-b\le(p)\le(b)$ and $q<-\frac{a}{2}$

$y=-\frac{a}{2}$

So the solutions are $(\frac{b}{2},\frac{bq}{2p}),(\frac{b}{2},-\frac{bq}{2p}),(\frac{pa}{2q},\frac{a}{2})(\frac{pa}{2q},-\frac{a}{2})$