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Get the equations of both lines going through $0$ which have a distance of 5 from the point $(1,7)$.

How to handle this problem? We have this formula:

If line $l$ is $ax+by=c$, distance $ P(x,y) $ to line $l$:

$ \dfrac{|ax+by-c|}{\sqrt{a^2+b^2}}$

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4 Answers 4

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Here is another approach, but I think this is more geometric than analytic geometry.

Let $Q(x,y)$ be the leg of the perpendicular from $P$ onto $l$.

Then, $PQ=5$ thus

$$(x-1)^2+(y-7)^2=25$$

Also, by the Pytagorean Theorem we have

$$PQ^2 +QO^2=PO^2$$

or

$$25+x^2+y^2=50$$

The two equations yield

$$x^2+y^2=25$$ $$2x+7y=50$$

solving for $x$ in the second equation and plugging in the first gives a quadratic, so the rest is easy.

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Find the circle with radius $5$ and center $(1,7)$ and then find the lines passing through the origin and which is a tangent to the circle by using the formula $y = mx+r\cdot\sqrt{m^2+ 1}$ and as it passes through tho origin $r.\sqrt{m^2+1}=0$ so your line now becomes $y=mx$ if distance of this line from $(1,7)$ is $5$, then you get a quadratic in $m$.

Solving this you can find out the values of $m$ and substituting in the line equation, we get the required equation.

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@Amzoti .... This is an answer. no tags needed because the question is already tagged. –  AvatarOfChronos Jan 11 '13 at 19:22

Here the lines pass through the origin $(0,0),$ so $c=0$

$$\frac{|a\cdot 1+b \cdot 7|}{\sqrt{a^2+b^2}}=5$$

Squaring, $(a+7b)^2=25(a^2+b^2)$

$\implies 24a^2-14ab-24b^2=0$

Solving for $a, a=\frac{4b}3$ or $-\frac{3b}4$

If $a=\frac{4b}3, \frac{4b}3x+by=0\implies 3y+4x=0$ as $b\ne 0$

If $a=-\frac{3b}4, -\frac{3b}4x+by=0\implies -3x+4y=0$ as $b\ne 0$


Here the lines pass through the origin $(0,0),$

the equation of the required line can be $y=mx$ where the gradient $m$ is finite, as the line is not the Y-axis.

So the distance of $(1,7)$ from $mx-y=0$ is $$\frac{|m\cdot 1-1\cdot 7|}{\sqrt{m^2+1}}=5$$

Squaring we get, $(m-7)^2=25(m^2+1)$

$24m^2+14m-24=0\implies m=-\frac 4 3$ or $\frac 3 4$

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I understand what you're doing up until 'solving for a'.. how did you do that? –  JohnPhteven Oct 13 '12 at 15:25
    
@ZafarS, it's a quadratic equation of $a$. You can put $a=b\cdot c$ and eliminate $b$ as $b\ne 0$ –  lab bhattacharjee Oct 13 '12 at 15:27
    
Might be easier to work with $y=mx$. Note that the line cannot be vertical, thus you can work with the slope formula... Then it is easier to work with only 1 variable... –  N. S. Oct 13 '12 at 15:32
    
My plan was originally to find the line through the points $(0,0)$ and $(1,7)$.. –  JohnPhteven Oct 13 '12 at 15:33
    
@N.S., I just wanted to keep the solution generic. Clearly, the line is not the Y-axis, so we can go with $y=mx$ –  lab bhattacharjee Oct 13 '12 at 15:38

First find the equation of the circle of radius 5 around the point. Then the lines you are looking for are the tangents of this circle which pass through 0.

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