Question

How do you find the equation of the line passing through the points $(1,4)$ and $(-2,6)$?

Answer 1

Step-1: Find slope of line joining the given two points:

Given points are $(1,4)$ and $(-2,6)$.

Slope of a line is $\mathrm{m}=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}$

For, given points $(1,4)$ and $(-2,6)$, $({\mathrm{x}}_1,{\mathrm{y}}_1)=(1,4)$ and $({\mathrm{x}}_2,{\mathrm{y}}_2)=(-2,6)$

So, $\mathrm{m}=\frac{6-4}{-2-1}$

$\Rightarrow$$\mathrm{m}=-\frac{2}{3}$

Step-2: Find the required equation of the line:

The equation of a line in slope-intercept form is $\mathrm{y}=\mathrm{mx}+\mathrm{b}$, where $\mathrm{m}$is the slope and $\mathrm{b}$ is $\mathrm{y}-$intercept.

Substitute $\mathrm{m}=-\frac{2}{3}$in slope-intercept form of the line;

$\Rightarrow$$\mathrm{y}=\left(-\frac{2}{3}\right)\mathrm{x}+\mathrm{b}$ $...\left(\mathrm{i}\right)$

Since, we have $({\mathrm{x}}_1,{\mathrm{y}}_1)=(1,4)$ , so line becomes;

$\Rightarrow$ $4=\left(-\frac{2}{3}\right)\left(1\right)+\mathrm{b}$

$\Rightarrow$$4+\frac{2}{3}=\mathrm{b}$

$\Rightarrow$ $\frac{14}{3}=\mathrm{b}$ $...\left(\mathrm{ii}\right)$

Substituting $\left(\mathrm{ii}\right)$ in $\left(\mathrm{i}\right)$, we get; $\mathrm{y}=\left(-\frac{2}{3}\right)\mathrm{x}+\frac{14}{3}$

Hence, the required equation of line is $\mathrm{y}=\frac{-2}{3}\mathrm{x}+\frac{14}{3}$.