Question

Can you help me solve thus?

Answer 1

First, we check that the three points are collinear.

Given three points (x1,y1), (x2,y2), and (x3,y3), then the three points are collinear if the following condition is satisfied.

`(y_2-y_1)/(x_2-x_1)=(y_3-y_1)/(x_3-x_1)`

In our case, we can set

(x1,y1) =(-2,0), (x2,y2) = ( 4, 18)and (x3,y3) = (5, 21)

`\begin{gathered} \text{ Then, we have} \\ (y_2-y_1)/(x_2-x_1)=(18-0)/(4-(-2))=(18)/(4+2)=(18)/(6)=3 \end{gathered}`

And

`(y_3-y_1)/(x_3-x_1)=(21-0)/(5-(-2))=(21)/(5+2)=(21)/(7)=3`

Hence

`(y_2-y_1)/(x_2-x_1)=(y_3-y_1)/(x_3-x_1)`

Therefore, the points (-2,0), ( 4, 18), and (5, 21) are collinear.

Given two points (x1,y1) and (x2,y2) on the Cartesian plane, then the gradient, m, of the line that passes through them is given by

`m=(y_2-y_1)/(x_2-x_1)`

Then, the gradient of the line passing through (-2,0), ( 4, 18) is 3

Given a point (x1,y1) on the Cartesian plane, and a line that passes through the point with gradient, m, then the equation of the line is given by

`y-y_1=m(x-x_1)`

Therefore,

`y-0=3(x-(-2)=3(x+2)`

That is the equation is

`y=3x+6`