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,
[tex]\begin{gathered} m=\frac{y2-y1}{x2-x1} \\ m=\frac{-5-2}{2-4} \\ m=\frac{-7}{-2} \\ m=\frac{7}{2} \end{gathered}[/tex]
The point slope form of the equation of a line can be written as,
[tex]y-y1=m(x-x1)[/tex]
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).
[tex]\begin{gathered} y-2=\frac{7}{2}(x-4) \\ 2(y-2)=7(x-4) \\ 2y-2\times2=7x-4\times7 \\ 2y-4=7x-28 \\ 2y=7x-28+4 \\ 2y=7x-24 \\ y=\frac{7}{2}x-\frac{24}{2} \\ y=\frac{7}{2}x-12\text{ ----(1)} \end{gathered}[/tex]
The general equation of a straight line is,
[tex]y=mx+b\text{ ------(2)}[/tex]
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,
[tex]y=\frac{7}{2}x-12\text{ -----(3)}[/tex]
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.
[tex]\begin{gathered} y=\frac{7}{2}\times(-4)-12 \\ =-14-12 \\ =-26 \end{gathered}[/tex]
Since we got y=-26 instead of , (-4, -2) is not a point on the line.
