Here in my book they have shown interpolation using parametric equation of line but I am unable to figure out how do they found out t=|y1-ys|/y1-y2 , t=|AD|/|AB| , t=|CE|/CB ,t=|EP|/|DE| here is the pic enter image description here

enter image description here

  • $\begingroup$ I prefer to think that linear interpolation is a weighted average with linked weights so that $p_i = t p_1 + (1 - t) p_2$ because this is easier in vector form and also motivated by vector drawings. Obviously its the same thing but expressed differently. $\endgroup$
    – joojaa
    Jan 15 '21 at 22:24

Since this is linear interpolation it boils down to solving a linear equation

$y = a + b t$

You can substitute the following $y = y_s$, $a = y_1$(intercept) and $b = y_2 - y_1$ (slope) to get the following

$$y_s = y_1 + t(y_2 - y_1)$$ $$y_s - y_1 = t(y_2 - y_1)$$ $$\frac{y_s - y_1}{y_2 - y_1} = t$$

The same thing happens for points where you can write

$$D = A + (B-A)t$$

$$ (D-A) = (B-A)\cdot t$$ $$ (D-A)^2 = ((B-A)\cdot t)^2$$ $$ \sqrt{(D-A)^2} = \sqrt{((B-A)\cdot t)^2}$$ $$ \sqrt{(D-A)^2} = \sqrt{(B-A)^2 \cdot t^2}$$ $$ \sqrt{(D-A)^2} = \sqrt{(B-A)^2} \cdot\sqrt{ t^2}$$ $$ |D-A| = |B-A| \cdot t$$ $$ t = \frac{|D-A|}{|B-A|}$$

  • $\begingroup$ But how could the slope be y2- y1 slope is : change in y over change in x i.e ∆y/∆x $\endgroup$
    – anuj goyal
    Jan 13 '21 at 13:47
  • $\begingroup$ actually the slope in this case is ∆y/∆t $\endgroup$
    – Reynolds
    Jan 14 '21 at 11:13

I4=I1 ( 1-t ) + t.I2 ; from parametric equation of line

I4=I1 - I1.t + t. I2

I4=I1+ t (I2-I1) Let's take line AB apart from triangle

enter image description here below in the equation I multiply t2 to numerator an denominator


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