Distance of a Point from a Line

Given,

$$x+y=5$$ and $$x^2+y^2=13$$

Using

$$(x+y)^2=x^2+y^2+2xy,$$

$$25=13+2xy$$

$$xy=6$$

Thus,

$$x,y$$ are roots of $$t^2-5t+6=0$$

$$=(t-2)(t-3)$$

Hence intersection points are

$$\left(2,3\right),\qquad \left(3,2\right)$$

Equation of the line joining them:

$$x+y=5$$

Distance of $$\left(2,-1\right)$$ from this line is $$\frac{|2-1-5|}{\sqrt2}$$

$$=\frac4{\sqrt2}$$

Thus,

$$k=4$$

Rearrange the family of lines by grouping terms with $$\lambda$$:

$$(x + 2y - 5) + \lambda(3x + 7y - 17) = 0$$

This is in the form $$L_1 + \lambda L_2 = 0$$. To find the fixed point $$P$$, solve the system:

$$x + 2y = 5$$

$$3x + 7y = 17$$

Multiplying the first equation by 3 and subtracting gives $$y = 2$$, which means $$x = 1$$. So, $$P = (1, 2)$$.

The line from this family that is farthest from the origin is the line perpendicular to the segment $$OP$$ (where $$O$$ is the origin $$(0,0)$$).

Slope of $$OP = \frac{2 - 0}{1 - 0} = 2$$. Therefore, the slope of line $$L$$ is $$-\frac{1}{2}$$.

Equation of line $$L$$ passing through $$P(1, 2)$$:

$$y - 2 = -\frac{1}{2}(x - 1) \implies x + 2y - 5 = 0$$

$$d = \frac{|3 + 2(6) - 5|}{\sqrt{1^2 + 2^2}} = \frac{|3 + 12 - 5|}{\sqrt{5}} = \frac{10}{\sqrt{5}}$$

$$d^2 = \frac{100}{5} = 20$$