Find point of intersection of two gradients

Question

I have four points $A$, $B$, $C$, $D$ on the Cartesian plane, with coordinates $(x_a,y_a)$, $(x_b,y_b)$, $(x_c,y_c)$, $(x_d,y_d)$. The lines $AB$, $BC$, $CD$ all have a gradient greater than or equal to $0$. Thus the implied gradient at B is the gradient $AC$, and the gradient at $C$ is the gradient $BD$.

How do I find the point of intersection of the gradients at $B$ and $C$? I spent a while faffing around with trig functions but ended up in endless recursive calculus-style loops. I suspect I’m missing something embassassingly obvious - perhaps because I’m on holiday and excessively relaxed :-)

Would appreciate some pointers, even though it’s high school level or less.

Answer 1

Write the two lines as functions of x: $$y_{AC}(x) = y_B + (x - x_B)\frac{y_C - y_A}{x_C - x_A}$$ and $$y_{BD}(x) = y_C + (x - x_C)\frac{y_D - y_B}{x_D - x_B}$$ The intersection occurs at $x$, $$y_{AC}(x) = y_{BD}(x)$$ If you expand all the terms, and move the coefficients (multipliers) of $x$ to the left side, and the others to the right side, you can divide both sides by the coefficients of $x$ to get the $x$ coordinate of the intersection point.

Insert into $y_{BD}(x)$ or $y_{AC}(x)$ to get the $y$ coordinate; both will give the same result.

Answer 2

You find the line equation of two lines then you solve these simultaneously. For the line at B, $$y-B_y=m_1(x-B_x)$$ and $$m_1=\frac{C_y-A_y}{C_x-A_x}$$ For the line at C $$y-C_y=m_2(x-C_x)$$ and $$m_2=\frac{D_y-B_y}{D_x-B_x}$$ Just plug in numbers for the points and then solve the two equation simultaneously for $x$ and $y$. Caveat: as stated in question, slopes are positive.