Line, points, distances and a parabola with maximum vertex

Text

Given the points $A(3;1)$, $B(-4;5)$ and the line $s:2x-3y+1=0$:

• (a) write $s$ in explicit form and draw it;
• (b) find the line through $A$ and $B$ and its intersection with $s$;
• (c) determine the inclination angle of $s$ with the $x$-axis;
• (d) find the line through $A$ parallel to $s$;
• (e) find the line through $B$ perpendicular to $s$;
• (f) find the distance from $A$ to $s$;
• (g) find the distance from $A$ to $B$;
• (h) draw the parabola $y=-\dfrac{x^2}{2}-4x-3$ and prove that $B$ is its maximum and that $A$ is not on the graph.

Solution

(a) $s:\ y=\dfrac23 x+\dfrac13$. (b) Line $AB$: slope $-\dfrac47$, so $y=-\dfrac47 x+\dfrac{19}{7}$. Intersection with $s$: $x=\dfrac{25}{13},\ y=\dfrac{21}{13}$, i.e. $\left(\dfrac{25}{13};\dfrac{21}{13}\right)\approx(1{,}92;1{,}62)$. (c) $\tan\alpha=\dfrac23\Rightarrow\alpha\approx33{,}69^\circ$. (d) Parallel through $A(3;1)$: $y=\dfrac23 x-1$ (i.e. $2x-3y-3=0$). (e) Perpendicular through $B(-4;5)$: $y=-\dfrac32 x-1$. (f) $d(A,s)=\dfrac{|6-3+1|}{\sqrt{13}}=\dfrac{4}{\sqrt{13}}=\dfrac{4\sqrt{13}}{13}\approx1{,}11$. (g) $d(A,B)=\sqrt{49+16}=\sqrt{65}\approx8{,}06$. (h) Parabola: $x_V=-4$, $y_V=5$: vertex $(-4;5)=B$. Since $a=-\dfrac12<0$, $B$ is the maximum. For $A$: $y(3)=-19{,}5\neq1$, so $A$ is not on the graph. $\boxed{AB\cap s=\left(\tfrac{25}{13};\tfrac{21}{13}\right);\ \alpha\approx33{,}69^\circ;\ d(A,s)=\tfrac{4\sqrt{13}}{13};\ d(A,B)=\sqrt{65};\ V=B(-4;5)\ \text{maximum}.}$