Respuesta :
Answer:
[tex]y+2=-\frac{3}{2}(x-2)[/tex]
Step-by-step explanation:
Given:
Two points on a line are given as:
(0, 1) and (2, -2)
We know that the slope of a line passing through points [tex](x_1,y_1)\ and\ (x_2,y_2)[/tex] is given as:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]
Plug in [tex]x_1=0,y_1=1,x_2=2,y_2=-2[/tex]. This gives,
[tex]m=\frac{-2-1}{2-0}\\\\m=\frac{-3}{2}=-\frac{3}{2}[/tex]
The equation of a line with slope 'm' and points [tex](x_1,y_1)\ and\ (x_2,y_2)[/tex] is given as:
[tex]y-y_1=m(x-x_1)\\\\or\\\\y-y_2=m(x-x_2)[/tex]
Now, on plugging in the second point [tex](x_2,y_2)=(2,-2)[/tex] we get:
[tex]y-(-2)=-\frac{3}{2}(x-2)\\\\y+2=-\frac{3}{2}(x-2)[/tex]
Therefore, the first option is correct which is given as:
[tex]y+2=-\frac{3}{2}(x-2)[/tex]
The equation that passes through the given orderd pairs is required.
The correct option is [tex]y+2=-\dfrac{3}{2}(x-2)[/tex]
The ordered pairs are
[tex](0,1),(2,-2)[/tex]
Let us substitute in each equation
[tex](0,1)[/tex]
[tex]y+2=-\dfrac{3}{2}(x-2)\\\Rightarrow 1+2=-\dfrac{3}{2}(0-2)\\\Rightarrow 3=3[/tex]
[tex](2,-2)[/tex]
[tex]-2+2=-\dfrac{3}{2}(2-2)\\\Rightarrow 0=0[/tex]
So, equation [tex]y+2=-\dfrac{3}{2}(x-2)[/tex] will pass through the given points.
[tex](0,1)[/tex]
[tex]y-3=\dfrac{3}{2}(x+1)\\\Rightarrow 1-3=\dfrac{3}{2}(1+1)\\\Rightarrow -2\neq3[/tex]
[tex](0,1)[/tex]
[tex]y+1=-\dfrac{2}{3}(x-3)\\\Rightarrow 1+1=-\dfrac{2}{3}(1-3)\\\Rightarrow 2\neq \dfrac{4}{3}[/tex]
[tex](0,1)[/tex]
[tex]y-4=-\dfrac{2}{3}(x+2)\\\Rightarrow 1-4=-\dfrac{2}{3}(0+2)\\\Rightarrow -3\neq -\dfrac{4}{3}[/tex]
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