Equation of straight line with given points

Question

I have Points $M_1(-1,1)$ and $M_2(2,3)$. From applied formula: $$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1).$$

Solving that gives: $$y-1=\frac{3-1}{2-(-1)}(x-(-1)).$$

which when solved becomes: $2x - 3y + 5 = 0$.

I cannot understand how I get that last result. Can you please explain

Answer 1

You can proceed as -

$$y-1=\frac{3-1}{2-(-1)}(x-(-1))$$

$$y-1=\frac{2}{3}(x+1).$$

$$3(y-1)=2(x+1)$$

$$3y-3=2x+2$$

$$2x-3y+5=0$$

Answer 2

you can compute the slope at first $$m=\frac{3-1}{2+1}$$ and we get by the Ansatz $$y=mx+n$$ with the $$m=\frac{2}{3}$$ the equation $$y=\frac{2}{3}x+n$$ plugging $$x=-1,y=1$$ in the given equation we have $$1=-\frac{2}{3}+n$$ this gives: $$n=\frac{5}{3}$$ and our equation will be $$y=\frac{2}{3}x+\frac{5}{3}$$ multiplying by $5$ we get $$3y=2x+5$$

Answer 3

Just simplify to get $$y-1=\frac{2}{3}(x+1)$$ which means that $$3(y-1)=2(x+1).$$ And so, we get $$3y-3=2x+2.$$ Transpose all the terms at the right side, we get $$0=2x+2-3y+3.$$ Finally, $$2x-3y+5=0.$$