Question

3/5 3) Find the equation of the line passing through the points (1, -2) and (-2, 7). Write the equation in slop-intercept from. Oy = 3x - 5 Oy= 3x + 13 Oy=-3x + 1 Oy=-3x - 2 Previous Next Submit arch O

Answer 1

hello

to solve this question we would first of all find the slope of the line

`\begin{gathered} slope(m)=(y_2-y_1)/(x_2-x_1) \\ \end{gathered}`
`\begin{gathered} y_1=-2 \\ x_1=1 \\ y_2=7 \\ x_2=-2 \end{gathered}`

we can now substitute this into the equation

`\begin{gathered} m=(y_2-y_1)/(x_2-x_1) \\ m=(7-(-2))/(-2-1) \\ m=(7+2)/(-3)=(9)/(-3)=-3 \end{gathered}`

the slope of the line is -3

we can write out how the equation looks

`\begin{gathered} y=mx+b \\ y=-3x+b \\ b=\text{intercept} \end{gathered}`

we can use one of the points to solve for b

`\begin{gathered} y=-3x+b \\ u\sin g\text{ the first point} \\ (1,-2) \\ -2=-3(1)+b \\ -2=-3+b \\ b=3-2 \\ b=1 \end{gathered}`

we can now rewrite our equation with slope and intercept

`\begin{gathered} y=mx+b \\ y=-3x+1 \end{gathered}`

from the calculation above, the equation of the line is given as y = -3x + 1