Line through two points: equation, perpendicularity and intersection

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A line $g$ passes through the points $(0;2)$ and $(2;3)$.

• (a) find the equation of $g$ in explicit and implicit form;
• (b) show that $g$ is perpendicular to the line $d:2x+y-4=0$;
• (c) find the intersection of $g$ and $d$.

Solution

(a) Slope $m_g=\dfrac{3-2}{2-0}=\tfrac12$; with intercept $2$: $y=\tfrac12x+2$ (explicit form), i.e. $x-2y+4=0$ (implicit form).

(b) Line $d$ has slope $-2$. Since $m_g\cdot m_d=\tfrac12\cdot(-2)=-1$, the two lines are perpendicular.

(c) Equating: $\tfrac12x+2=-2x+4\Rightarrow \tfrac52x=2\Rightarrow x=\tfrac45$; then $y=\tfrac12\cdot\tfrac45+2=\tfrac{12}{5}$. Intersection $\left(\tfrac45;\tfrac{12}{5}\right)=(0{,}8;2{,}4)$.

$\boxed{g:\ y=\tfrac12x+2,\quad P\left(\tfrac45;\tfrac{12}{5}\right)=(0{,}8;2{,}4)}$