Finding the equation of a line whose segment is intercepted between axes

Question

The question is: Find the equation of a line passing through $(-2, 5)$ and whose segment intercepted between axes in the 2nd quadrant is $7\sqrt{2}$

I have two graphs in mind but I don't know which one is correct. The first graph is a line from point $(-2, 5)$ to the origin since the questions says "between axes". The second graph is a line from $(-2, 5)$ to the $x$ and $y$ axis. Which one is correct? I'm going to start with distance formula, right?

Answer 1

What we have is a line with positive slope, passing through the point $(-2, 5)$, with the length of the segment on this line from the point where it intersects the negative $x$-axis at $(x_0, 0)$ and the positive $y$-axis at $(0, y_0)$ equal to $7\sqrt 2$.

We know then, that $$x_0^2 + y_0^2 = (7\sqrt 2)^2 = 98\tag{1}$$

We also know that the distance from $(x_0, 0)$ to $(-2, 5)$ plus the distance from $(-2, 5)$ to the point $(0, y_0)$ is equal to $7\sqrt 2$: $$\sqrt {25 + (x_0+2)^2} + \sqrt{4+(y_0-5)^2} = 7\sqrt 2\tag{2}.$$

Solving $(1)$ and $(2)$ simultaneously gives us the real solutions $x_0 = -7$, and $y_0 = 7$. With these values, we must have that the slope of each of these "segments" comprising the entire length are equal: $$m = \frac{5-0}{-2 - x_0} = \frac{5-y_0}{-2-0}= 1$$

Now, given your point on the line $(-2, 5)$ and the slope of $m = 1$, we can construct the equation of the line: $$y - 5 = x + 2 \iff y = x+7$$

Answer 2

If it went through the origin then its length would be $\sqrt{5^2+2^2}=\sqrt{29}\lt7\sqrt{2}$. The question states that this line segment is intercepted between the axes. I would therefore conclude that is must intercept BOTH the x- and y-axes.

As discussed in comments below, lets assume the lines crosses the x-axes at $(x_1,0)$ and the y-axes at $(0,y_1)$. You are told that the length of the line segment is $7\sqrt{2}$, therefore:$$x_1^2+y_1^2=(7\sqrt{2})^2=98$$ You are also told that the line passes through $(-2,5)$. Therefore:

Slope from $(0,y_1)$ to $(x_1,0)$ is:$$\frac{y_1}{x_1}\tag{1}$$Slope from $(0,y_1)$ to $(-2,5)$ is:$$\frac{y_1-5}{0-(-2)}=\frac{y_1-5}{2}\tag{2}$$Slope from $(-2,5)$ to $(x_1,0)$ is:$$\frac{5-0}{-2-x_1}=-\frac{5}{2+x_1}\tag{3}$$All 3 slopes must be equal. Use these to solve for $x_1$ and $y_1$.

Answer 3

Let the eqn of the line be $ y=mx+c$, point on the X-axis be $(-x,0)$, point on the Y-axis be $(0,y)$

When $y=0$ (i.e. on X-axis) we get $ 0=m(-x)+c\Rightarrow mx=c$...(I)
When $x=0$ (i.e. on Y-axis) we get $y=c$...(II)
For the point $(-2,5)$ we get $5=-2m+c$...(III)

From (III) $c=5+2m$
$ \therefore $ From (II) $ y=5+2m$
From (I) $mx=5+2m \Rightarrow x=\frac{5+2m}{m}$

Substitute $x$ and $y$ in terms of $m$ in the eqn $x^2+y^2=98$

$$(5+2m)^2+\left( \frac{5+2m}{m}\right)^2=98$$

Upon first factorisation we get $(m-1)(4m^3+24m^2-45m-25)=0$

$\therefore m=1$ (since $m$ is real and positive and a line can not have more than one slope, it agrees with our problem and also I don't know what can be done to the polynomial in the second bracket)

Substituting $m=1$ and $(-2,5)$ in $y=mx+c$ we get $c=7$

So the equation of the line is $y=x+7$