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
    Commented Jan 15, 2021 at 22:24

2 Answers 2


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
    Commented Jan 13, 2021 at 13:47
  • $\begingroup$ actually the slope in this case is ∆y/∆t $\endgroup$
    – Reynolds
    Commented Jan 14, 2021 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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.