The line equation that passes through the given points is
![3x-y + 1 = 0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/88gndhkrd14d8jyoofic80j0m8nyx0a5zj.png)
SOLUTION:
Given, two points are A(-1, -2) and B(3, 10).
We need to find the line equation that passes through the given two points. We know that, general equation of a line passing through two points
is given by
![(y-y_(1))/(x-x_(1))=(y_(2)-y_(1))/(x_(2)-x_(1))](https://img.qammunity.org/2020/formulas/mathematics/middle-school/shmxec31s7q7hfk5weaplmg4e75ctogh12.png)
This can be written as,
![y-y_(1)=\left((y_(2)-y_(1))/(x_(2)-x_(1))\right)\left(x-x_(1)\right) \rightarrow(1)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/p442h6gpcjujes8rfje02nnvrsuhfm7rxb.png)
here, in our problem
![x_1 = - 1; y_1 = - 2; x_2 = 3 \text { and } y_2 = 10.](https://img.qammunity.org/2020/formulas/mathematics/middle-school/4pw1rga20nqmom4yd8hn7r5wgnar4njz3o.png)
Now substitute the values in (1)
![\begin{array}{l}{y-(-2)=\left((10-(-2))/(3-(-1))\right)(x-(-1))} \\\\ {\Rightarrow y+2=(10+2)/(3+1)(x+1)} \\\\ {\Rightarrow y+2=(12)/(4)(x+1)} \\\\ {\Rightarrow y+2=3(x+1)} \\\\ {\Rightarrow y+2=3 x+3} \\\\ {\Rightarrow 3 x-y+3-2=0} \\\\ {\Rightarrow 3 x-y+1=0}\end{array}](https://img.qammunity.org/2020/formulas/mathematics/middle-school/s7ys4cpp5w9ujo539s60h7dht6fuh0pd6p.png)
Hence, the line equation that passes through the given points is
![3x-y + 1 = 0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/88gndhkrd14d8jyoofic80j0m8nyx0a5zj.png)