Problem:
Consider a family of straight lines $(x+y)+\lambda(2x-y+1)=0$. Find the equation of the straight line belonging to this family which is farthest from $(1,-3)$.
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Any help with this problem would be really appreciated!
[Math] Family of Lines
analytic geometrygeometry
Related Solutions
$$x^2 - y^2 + 2y = 1\iff x^2 = y^2 -2y+1 = (y-1)^2 \iff y-1 = \pm x \iff y = 1\pm x$$
Thus the two intersecting lines are given by $y = 1+x$ and $y = 1-x$. The point of intersection is given by $(0, 1)$. One bisector is the y-axis: the line $x=0$. The other bisector is given by $y = 1$.
Can you take it from here?
First, note that all the lines that can be represented by your equation pass through the point (-2,0). Obviously it follows that the foot of the perpendicular to the line from (2,3) should be (-2,0)
Best Answer
Think of it geometrically. Since the family of lines all pass through $I\left(-\frac{1}{3}, \frac{1}{3}\right)$, the largest distance from any point $A$ to a line passing through $I$ will be the length $IA$. Therefore, the problem becomes:
Solution
The slope of the line passing through $IA$ is
You want the slope of the line perpendicular to this, which is
Combined with the known point, the equation is
Edit: Reasoning for perpendicularity
Let $H$ be the intersection of a perpendicular line drawn from $A$. We have $AH$ is the distance from $A$ to the line.
Suppose $IA$ makes an angle $\theta$ with the line in question, it follows that $AH = IA\sin\theta \le IA$. Thus $AH = IA$ when $\theta = 90^\circ$ or $H \equiv I$