Calculus 10th Edition

(a)$$y=- \frac{2}{3}x+\frac{16}{3}$$ (b)$$y=x+2$$ (c)$$y=- \frac{3}{4}x+\frac{11}{2}$$ (d)$$y=4$$
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; $$y_1=mx_1+d \quad \Rightarrow \quad 4= 0 \cdot 2+d \quad \Rightarrow \quad d=4$$$$\Rightarrow y=4$$