+5478 First, you would need to find the slope of the line.
Use the formula $${\mathtt{m}} = {\frac{\left({\mathtt{x1}}{\mathtt{\,-\,}}{\mathtt{x2}}\right)}{\left({\mathtt{y1}}{\mathtt{\,-\,}}{\mathtt{y2}}\right)}}$$
Plug in the points:
$${\mathtt{m}} = {\frac{\left({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}{\left({\mathtt{5}}{\mathtt{\,-\,}}{\mathtt{4}}\right)}}$$
m=3
Now use the slope and a point to write the equation in point-slope form:
$$\left({\mathtt{y}}{\mathtt{\,-\,}}{\mathtt{y1}}\right) = {m}{\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{x1}}\right)}$$
$${\mathtt{y}}{\mathtt{\,-\,}}{\mathtt{3}} = {\mathtt{3}}{\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{4}}\right)$$
Simplify it:
y= 3x-12+3
y= 3x-9
+5478 First, you would need to find the slope of the line.
Use the formula $${\mathtt{m}} = {\frac{\left({\mathtt{x1}}{\mathtt{\,-\,}}{\mathtt{x2}}\right)}{\left({\mathtt{y1}}{\mathtt{\,-\,}}{\mathtt{y2}}\right)}}$$
Plug in the points:
$${\mathtt{m}} = {\frac{\left({\mathtt{6}}{\mathtt{\,-\,}}{\mathtt{3}}\right)}{\left({\mathtt{5}}{\mathtt{\,-\,}}{\mathtt{4}}\right)}}$$
m=3
Now use the slope and a point to write the equation in point-slope form:
$$\left({\mathtt{y}}{\mathtt{\,-\,}}{\mathtt{y1}}\right) = {m}{\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{x1}}\right)}$$
$${\mathtt{y}}{\mathtt{\,-\,}}{\mathtt{3}} = {\mathtt{3}}{\mathtt{\,\times\,}}\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{4}}\right)$$
Simplify it:
y= 3x-12+3
y= 3x-9
