Question

Show that the points $(1,4),(3,-2)$ and $(-3,16)$ are collinear. Find equation of the line through them.

Answer 1

Step1: Calculation of equation of the line passing through $(1,4),(3,-2)$.

The two-point form of the equation of a line passing through $\left(x_1,y_1\right)$and $\left(x_2,y_2\right)$ is given by the formula $\left(y-y_1\right)=\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\left(x-x_1\right)$.

Thus, the equation of a line passing through $(1,4),(3,-2)$ is given by:

$$\begin{gathered} \left(y-4\right)=\frac{\left(-2-4\right)}{\left(3-1\right)}\left(x-1\right) \\ \Rightarrow \left(y-4\right)=\frac{-6}{2}\left(x-1\right) \\ \Rightarrow \left(y-4\right)=-3\left(x-1\right) \\ \Rightarrow \left(y-4\right)=-3x+3 \\ \Rightarrow y-4+3x-3=0(adding3x-3tobothsides) \\ \therefore 3x+y-7=0 \end{gathered}$$

Step2: Verification of collinearity of points $(1,4),(3,-2)$ and $(-3,16)$.

Substitute the point $(-3,16)$ in equation $3x+y-7=0$ and check whether the point satisfies the equation or not.

$\begin{array}{rcl}3(-3)+16-7& =& 0\\ -9+9& =& 0\\ 0& =& 0\end{array}$

Since, the the point $(-3,16)$ satisfies the equation $3x+y-7=0$; Thus the points $(1,4),(3,-2)$ and $(-3,16)$ are collinear.

The equation of line through these points is $3x+y-7=0$.

Hence, the equation of line through the collinear points $(1,4),(3,-2)$ and $(-3,16)$ is $3x+y-7=0$.