How to draw a line of intersection of a line at an angle?

Question

I have a line called $AB$ with the points $A (40, 40)$ and $B (280, 40)$. I have an angle $D\simeq 48^\circ$.

I want to draw a line $CE$ from the center of the line $AB$, which is $C (160, 40)$ with the given angle $D$.

Something like,

I have the line $AB$, the point $C$, and angle $D$. How can I find the point $E$, which is anywhere on the line of intersection with the given values?

In my analysis, I need to find out the slope of the new line first and then we need to get points of the line.

If that is right, how can I get the slope of the new line, which is intersecting given line?

If wrong what would be the right way?

Thanks.

Answer 1

I have made some calculation, hope this works.

Given that, $A(40,40)$ and $B(280,40)$ endpoints of the line $AB$, $C(160,40)$ is the midpoint of $AB$. Angle has given $D=48^\circ$ then the required line that passes through $C$ having slope $\tan(\pi-48^\circ)$ is, $$y-40=\tan(\pi-48^\circ)(x-160)$$ $$\text{i.e. }y-40=-\tan(48^\circ)(x-160)$$ any point on this line can be treated as $E$.

Answer 2

If point $E$ lies $r$ units away from $C$ (This distance isn't perpendicular distance) such that $CE=r$ then, co-ordinates of point $E$ will be

$(160-rcos\theta , 40+rsin\theta)\tag{1} $ where $\theta =48^{\circ}$

now, the question is what if line $AB$ isn't horizontal(as in your question) but

inclined at an angle $\phi$ with x-axis and C is also not a midpoint but a point which divides line $AB$ in ratio $m:n$ such that $AC=\dfrac{ml}{m+n}$ and $CB=\dfrac{nl}{m+n}$ where $l$ is length of line segment AB

then,

co-ordinates of $D$ become $\bigg(x_{A}+\dfrac{ml}{m+n}cos\phi \bigg),\bigg(y_{A}+\dfrac{ml}{m+n}sin\phi \bigg)$ here $(x_{A},y_{A})$ are co-ordinates of point A.

and, finally co-ordinates of E become $\bigg(x_{A}+\dfrac{ml}{m+n}cos\phi -rcos(\theta-\phi) \bigg),\bigg(y_{A}+\dfrac{ml}{m+n}sin\phi +rsin(\theta-\phi) \bigg)$

you can check put $\phi=0^{\circ} $ and $\theta=48^{\circ}$ alongwith co-ordinates of point A you get above equation $(1)$

i have used parametric form of line not slope point form

Answer 3

From given data input

$$\dfrac{y-40}{x-160}=\tan(180^{\circ}-48^\circ)$$

Length of $ CE $ does not matter because an angle does not depend on length of either of its two arms. Any or all lengths of $CE$ are already included in the above slope equation of straight line.

For construction using a protractor of any size, mark $48^{\circ} $ with $C$ as center in the clockwise direction from $AC$ and and join $EC$.