Question

In ℝ2, you are given the points A(−14,15) and X(13,−17). Find t such that the point C(−12,t) lies on the line through A and X. answer as a ratio not as a decimal

Answer 1

Answer:

The equation of the line segment between points A and X is

$sA + (1-s)X, s\in[0,1]$

Then we need that the point C satisfy:

$\left[\begin{array}{c}-12\\t\end{array}\right] =\left[\begin{array}{c}-14s\\15s\end{array}\right]+\left[\begin{array}{c}13(1-s)\\-17(1-s)\end{array}\right]$.

This implies that

$-12=-14s-13s+13\\s=\frac{25}{27}$

We replace the value of s in the equation we get from the second component

$t=15s+17s-17\\t=32\frac{25}{27}-17\\t=\frac{341}{27}$

Then the point $C=(-12,\frac{341}{27})$ lies on the line through A and X.