Figure 1 shows two points A and B with a straight line drawn through them.

(a)
(i) As is evident from the figure $$A=(-3,5),\qquad B=(2,-4) $$
(ii) We know that the gradient is the rise from A to B divided by the run from A to B, i.e. change in y-coordinate from A to B divided by the change in x-coordinate from A to B. So we have $$m=\frac{(-4)-5}{2-(-3)}=\frac{-9}{5}=-1.8 $$ 

(b) 
(i) The equation of the line passing through $(x_1,y_1)$ with slope $m$ is $$y-y_1 =m(x-x_1) $$ So in this case we have $$y-4=\frac{2}{5}(x-(-3)) \implies y=\frac{2}{5}(x+3)+4=\frac{2}{5}x+\frac{26}{5}$$ 

(ii) The point $(-2,5)$ is on the line if and only if it satisfies the equation of the line. For $x=-2$ and $y=5$ we have $$\frac{2}{5}x+\frac{26}{5}=\frac{2}{5}(-2)+\frac{26}{5}=\frac{-4+26}{5}=\frac{22}{5}=4.4\ne 5=y $$ So the point does not satisfy the equation of the line and thus it does not line on the line. 

(iii) The x-intercept is x-coordinate of the intersection of the line with the x-axis, and the x-axis is given by the equation $y=0$. Setting $y=0$ in the equation of the line gives    $$ 0=\frac{2}{5}x+\frac{26}{5} \implies \frac{2}{5}x=-\frac{26}{5}\implies x= -13$$ So the x-intercept is $-13$ corresponding to the intersection point $(-13,0)$ with the x-axis.