Find the equation of a line whose distance from a point is given

Question

The question is: If the distance of a point $(1, 4)$ from a line passing through the intersection of the lines $x-2y+3=0$ and $x-y-5=0$ is $4$ units. Find its equation.

Answer 1

hint.....The point of intersection is $(13,8)$. A line of gradient $m$ through this point is $$y-8=m(x-13)\Rightarrow y-mx+13m-8=0$$

Now you can use the formula for the distance $d$ from a point $(p,q)$ to the line $ax+by+c=0$ which is $$d=\left|\frac{ap+bq+c}{\sqrt{a^2+b^2}}\right|$$ You can now form and solve a quadratic equation to find the possible values of $m$

Answer 2

There are $2$ such lines.

The lines with equations $x-2y+3=0$ and $x-y-5=0$ define a pencil of lines: $$t(x-2y+3)+u(x-y-5)=0\iff (t+u)x-(2t+u)y+3t-5u=0$$ which pass through the intersection point of the given lines.

The distance from the point $(1,4)$ to one of these lines is given by: $$\frac{\lvert t+u-4(2t+u)+3t-5u\rvert}{\sqrt{(t+u)^2+(2t+u)^2}}=\frac{4\lvert t+2u\rvert}{\sqrt{5t^2+6tu+2u^2}},$$ and it is equal to $4$ if and only if $\;\lvert t+2u\rvert=\sqrt{5t^2+6tu+2u^2}$, i.e. $$(t+2u)^2=5t^2+6tu+2u^2\iff 4t^2+2tu-2u^2=0.$$ As this is a homogeneous quadratic equation, we solve it setting $\lambda=\dfrac tu$. We obtain the equation in $\lambda$: $\;2\lambda^2+\lambda-1=0$.

It has an integer root: $\lambda=-1$, so the other root, by Vieta's formulae is $\;\lambda=\dfrac12$. Thus the requested lines have equation:

• $y=8$ (for $\lambda=-1$, i.e. $u=-t$),
• $3x-4y-7=0$ (for $\lambda=\dfrac12$, i.e. $u=2t$).

Answer 3

Since we are seeking the equation(s) of the line(s) through an intersection point $ \ P \ $ that have a perpendicular distance $ \ 4 \ $ from the point $ \ C \ (1 \ , \ 4) \ \ , \ $ such a line will be tangent to the circle [marked in red in the graph above] of radius $ \ 4 \ $ centered on $ \ C \ \ , \ (x - 1)^2 + (y - 4)^2 \ = \ 4^2 \ \ . \ $ We have established that the intersection of the lines $ \ x - 2y + 3 \ = \ 0 \ $ and $ \ x - y - 5 \ = \ 0 \ $ [in green] is found from $ \ (y + 5) - 2y + 3 \ = \ 0 \ \Rightarrow \ y_P \ = \ 8 $ $ \Rightarrow \ x_P \ = 13 \ \ . \ $

In similar problems, we would be concerned with finding the slopes of the tangent lines more directly, but we have some very helpful information here. The radii of the circle centered on $ \ C \ $ [in orange] are perpendicular to the tangent lines at the tangent points, so we will be interested in the two right triangles with vertices at $ \ P \ , \ C \ $ and those tangent points. We find that the common hypotenuse [in violet] of those right triangles has a length given by the length of $ \ \overline{PC} \ \ , $ $$ s \ \ = \ \ \sqrt{ \ (13 \ - \ 1 )^2 \ + \ (8 \ - \ 4)^2} \ \ = \ \ \sqrt{144 \ + \ 16 } \ \ = \ \ \sqrt{160} \ \ = \ \ 4·\sqrt{10} \ \ . $$ The remaining leg of each right triangle thus has a length $ \ \sqrt{160 - 4^2} \ = \ \sqrt{144} \ = \ 12 \ \ . $

The tangent points are then at the intersection(s) of our circle centered on $ \ C \ $ and a circle of radius $ \ 12 \ $ centered on $ \ P \ \ , \ $ those points being obtained from $$ (x \ - \ 1)^2 \ + \ (y \ - \ 4)^2 \ \ = \ \ 16 \ \ \rightarrow \ \ x^2 \ - \ 2x \ + \ y^2 \ - \ 8y \ + \ 1 \ \ = \ \ 0 \ \ \ , $$ $$ (x \ - \ 13)^2 \ + \ (y \ - \ 8)^2 \ \ = \ \ 144 \ \ \rightarrow \ \ x^2 \ - \ 26x \ + \ y^2 \ - \ 16y \ + \ 89 \ \ = \ \ 0 $$

$$ \Rightarrow \ \ 2x \ + \ 8y \ \ = \ \ 1 \ \ , \ \ 26x \ + \ 16y \ \ = \ \ 89 \ \ \Rightarrow \ \ 3x \ + \ y \ \ = \ \ 11 \ \ . \ $$

The two tangent points lie on this line: inserting this into the equation of the circle centered on $ \ C \ $ yields $$ x^2 \ - \ 2x \ + \ (11 \ - \ 3x)^2 \ - \ 8·(11 \ - \ 3x) \ + \ 1 \ \ = \ \ 0 $$ $$ \Rightarrow \ \ 10·x^2 \ - \ 44·x \ + \ 34 \ \ = \ \ 2·(x \ - \ 1)·(5x \ - \ 17) \ \ = \ \ 0 \ \ . \ $$

Thus the coordinates of the tangent points are $$ \ x \ = \ 1 \ \ , \ \ x \ = \ \frac{17}{5} \ \ \Rightarrow \ \ y \ = \ 11 - 3·1 \ = \ 8 \ \ , \ \ y \ = \ 11 - 3·\frac{17}{5} \ = \ \frac45 \ \ . $$

With these tangent points now known to be $ \ (1 \ , \ 8) \ $ and $ \ \left(\frac{17}{5} \ , \ \frac45 \right) \ \ , \ $ we can determine the slopes of the lines passing through $ \ P \ $ as $$ m_1 \ \ = \ \ \frac{8 \ - \ 8}{13 \ - \ 1} \ \ = \ \ 0 \ \ \ , \ \ \ m_2 \ \ = \ \ \frac{8 \ - \ \frac45}{13 \ - \ \frac{17}{5}} \ \ = \ \ \frac{ \frac{36}{5}}{ \frac{48}{5}} \ \ = \ \ \frac34 \ \ . $$

The two lines [in blue] we have sought are therefore $ \ y - 8 \ = \ 0 \ \rightarrow \ \boxed{ \ y \ = \ 8 \ } \ \ $ and

$ \ y - 8 \ = \ \frac34·(x - 13) \ \rightarrow \ \boxed{ \ y \ = \ \frac34x \ - \ \frac74 \ } \ \ . $

[From a sketch of the geometrical situation, we might have concluded that, since the $ \ y-$coordinate of $ \ P \ $ is the same as that of the point $ \ 4 \ $ units "above" $ \ C \ (1 \ , \ 4) \ \ , \ $ one of the lines of interest is horizontal. We would still need to work to find the second line.]