Answer
(a)$$y=- \frac{2}{3}x+\frac{16}{3}$$
(b)$$y=x+2$$
(c)$$y=- \frac{3}{4}x+\frac{11}{2}$$
(d)$$y=4$$
Work Step by Step
The general equation of a line is $y=mx+d$.
In this question, the lines pass through the point $(2,4)$.
To find equations of the lines satisfying the given conditions, one must find unknown parameters ($m$ and $d$) by applying the conditions in the general form of equation of line.
(a) hypothesis: $(x_1,y_1)=(2,4)$ and $m=- \frac{2}{3}$; $$y_1=mx_1+d \quad \Rightarrow \quad 4= - \frac {2}{3} \cdot 2+d \quad \Rightarrow \quad d=\frac{16}{3}$$$$\Rightarrow y=- \frac{2}{3}x+\frac{16}{3}$$
(b) hypothesis: $(x_1,y_1)=(2,4)$ and the line is perpendicular to the line $y=-x$, that is, $m= - \frac{1}{-1}=1$;
$$y_1=mx_1+d \quad \Rightarrow \quad 4= 1 \cdot 2+d \quad \Rightarrow \quad d=2 $$$$\Rightarrow y=x+2$$
(c) hypothesis: $(x_1,y_1)=(2,4)$ and $(x_2, y_2)=(6,1)$, that is, $m=\frac{y_2-y_1}{x_2-x_1}=\frac{1-4}{6-2}=-\frac{3}{4}$;
$$y_1=mx_1+d \quad \Rightarrow \quad 4= - \frac {3}{4} \cdot 2+d \quad \Rightarrow \quad d=\frac{11}{2}$$$$\Rightarrow y=- \frac{3}{4}x+\frac{11}{2}$$
(d) hypothesis: $(x_1,y_1)=(2,4)$ and the line is parallel to the x-axis ($y=0$), that is, $m=$0;
$$y_1=mx_1+d \quad \Rightarrow \quad 4= 0 \cdot 2+d \quad \Rightarrow \quad d=4 $$$$\Rightarrow y=4$$
