Given:
The coordinates of points is
(x1, y1)=(4, 2)
(x2, y2)=(2, -5).
The slope of the line passing through points (x1, y1)=(4, 2) and (x2, y2)=(2, -5) is,
![\begin{gathered} m=(y2-y1)/(x2-x1) \\ m=(-5-2)/(2-4) \\ m=(-7)/(-2) \\ m=(7)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/2d0dft0nkoipwhbtfdth2pzdy3jl9nux52.png)
The point slope form of the equation of a line can be written as,
![y-y1=m(x-x1)](https://img.qammunity.org/2023/formulas/mathematics/high-school/w8833wq43gzv073odoa76ku9bunhdf4cj7.png)
Put (x1, y1)=(4, 2) in the above equation to find the equation of a line with slope m=7/2 and passing through (4,2).
![\begin{gathered} y-2=(7)/(2)(x-4) \\ 2(y-2)=7(x-4) \\ 2y-2*2=7x-4*7 \\ 2y-4=7x-28 \\ 2y=7x-28+4 \\ 2y=7x-24 \\ y=(7)/(2)x-(24)/(2) \\ y=(7)/(2)x-12\text{ ----(1)} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/th3iqlbxfutrir147hhle2l69hpd0d47js.png)
The general equation of a straight line is,
![y=mx+b\text{ ------(2)}](https://img.qammunity.org/2023/formulas/mathematics/college/px78kymuy6vml9p1qxnpyzfagqc1pq7x2s.png)
Here, m is the slope of the line and b is the y intercept.
Comaparing equations (1) and (2), we get y intercept b=-12.
Therefore, the y intercept b=-12.
The equation of the line is,
![y=(7)/(2)x-12\text{ -----(3)}](https://img.qammunity.org/2023/formulas/mathematics/college/41e5q9iqoju67ccxwkzy14h7ula9x1cbkm.png)
We have to check if (-4, -2) is a point on the line.
For that put x=-4 in equation (3) and solve for y to check if we get y=-2.
![\begin{gathered} y=(7)/(2)*(-4)-12 \\ =-14-12 \\ =-26 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/46c9iy4tvqwob5wa1zlz8985pivvuwbako.png)
Since we got y=-26 instead of , (-4, -2) is not a point on the line.