Let say I have this diagram,
How to find the direction vector passing through the intersection point of two straight lines?
Update: new vector is the bisector of two lines and vector may be arbitrary.
[Math] Get the direction vector passing through the intersection point of two straight lines
calculusgeometrylinear algebravector-spaces
Related Solutions
The intersection of the two lines is not $(13,4)$. Plugging that point into the first line gives $8$ and into the second gives $4$ for the left hand side, not $0$, so you need to redo the solution for the point. The $x$ coordinate is correct, but not $y$. Then there are many lines through that point and you are to use the distance information to pick two of them. If the intersection point is $(a,b)$, the point slope formula would be $y-b=m(x-a)$ and you need to use your formula for the distance between $(1,4)$ and this line to pick $m$.
Added: I meant the line through $(13,8)$ of slope $m$, so $y-8=m(x-13)$ or $-mx+y+13m-8=0$. You need the distance from $(1,4)$ to this line to be $4$. Mathworld gives the distance from the point $(x_0,y_0)$ to the line $ax+by+c=0$ as $\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$ so you have $4=\frac{|-m+4+13m-8|}{\sqrt{m^2+1}}$. Moving the $\sqrt{m^2+1}$ to the other side and squaring, you should find two values for $m$.
Take the general form $Ax + By + C = 0$ and $A_2x + B_2y + C_2 = 0$
Then the point of intersection will be $(\frac{B*C_2 - B_2*C_1}{A*B_2-A_2*B} , \frac{C*A_2 - A_1*C_2}{A*B_2-A_2*B})$
Now take $B_2$ as zero because line is parallel to y-axis.
Thus the new point of intersection will be $(\frac{B*C_2}{-A_2*B} , \frac{C*A_2 - A_1*C_2}{-A_2*B})$ = $(\frac{-C_2}{A_2} , \frac{A_1*C_2 -C*A_2}{A_2*B})$
And to find the angle between those two lines.
Best Answer
Normalize the two given vectors, i.e. compute $$e_a:={a\over |a|}={1\over\sqrt{13}}(2,3)\ ,\quad e_b:={b\over|b|}={1\over\sqrt{5}}(2,1)\ .$$ Then the angle bisector will pass through the point $$p:=e_a+e_b=\Bigl({2\over\sqrt{13}}+{2\over\sqrt{5}},{3\over\sqrt{13}}+{1\over\sqrt{5}}\Bigr)\doteq(1.44913,1.27926)$$ whose argument is $\arctan(0.882782)=0.723221=41.44^\circ$.
The following figure shows the geometric intuition behind the above computation: