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Okay, here is the question:

A straight line makes on the coordinate axes positive intercepts whose sum is $7.$ If the line passes through the point $(-3,8),$ find it's equation.

I spent an hour in the afternoon. I'm usually quite comfortable solving sums like these, but this one just messed me up. I'm not getting that peace of mind till I see this solved. I tried a way out:

Let the required points be $Q(0,y)$ and $P(x,0)$.

$x + y = 7$ as sum of x intercept and y intercept is seven.

So, $x = 7-y$

Thus, we can write $P(7-y,0)$ and $Q(0,y)$.

The slope of $PR = \Large \frac{8}{y-10}$, and the slope of $QR = \Large \frac{8-y}{-3}$

But, since the points are collinear, they have the same slope.

The trouble starts then with a quadratic that's bad in the literal sense when I equate them.

The answer is $4x+3y = 12$, which is the correct answer. I would like to learn how to solve these kind of problems are so very interesting. Thanks!

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  • $\begingroup$ @HarshKumar: I have rejected all of your edits for the following reasons: they were performed on old, dormant questions, and they weren't improving anything. After taking a look at your activity, it became clear that you perform many small useless edits only in order to gain badges and those meagre +2 reputation for each edit. Please stop this right now, otherwise I shall have to report you to a moderator. $\endgroup$ – Alex M. Jan 1 '17 at 17:55
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It's not a good idea to use "x" and "y" as the intercepts if you are going to use them as the variables in the equation also! Lets call the intercepts $(x_0, 0)$ and $(0, y_0)$. Then $x_0+ y_0= 7$ or, of course, $y_0= 7- x_0$. Now there are a number of ways to go but the simplest is to use the fact that if the intercepts of a straight line are (a, 0) and (0, b), then the line has equation bx+ ay= ab. You can see that by noting that if x= a, y= 0 then b(a)+ a(0)= ab is true and if x= 0, y= b, then b(0)+ a(b)= ab is true. So we have $(7- x_0)x+ x_0y= x_0(7- x_0)$ Set x= -3,y= 8 and you have a quadratic equation for $x_0$.

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Say the equation of the straight line form in intercept form is $\frac{x}{a}+\frac{y}{b}=1$

Given:

  1. $a+b=7$
  2. $\frac{-3}{a}+\frac{8}{b}=1 \Rightarrow -3b+8a=ab$

From $1$ and $2$, we have that $-3(7-a)+8a=a(7-a) \Rightarrow-21+3a+8a=7a-a^2$

Therefore we have $a^2+4a-21=0 \Rightarrow (a+7)(a-3)=0$

But it is stated that $a,b>0$

So the required equation for the straight line is:

$$\frac{x}{3}+\frac{y}{4}=1$$

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  • $\begingroup$ @hardmath Thanks for pointing it out. Missed it while reading. Edited it. $\endgroup$ – SchrodingersCat Nov 15 '15 at 12:58
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I don't see your problem: your equation is $\displaystyle\frac{8}{y-10}=\frac{8-y}{-3}$, that is $y^2-18y+56=0$, and its solutions are $y=4$ and $y=14$. Of course the second one must be discarded because it yields $x=-7$.

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suppose straight line intercepts $x$ axis at $(x_0,0)$ and $y$ axis at $(0,y_0)$

since $x_0+y_0=7$

$y_0=7-x_0$

our line can be written as

$\frac{y-0}{x-x_0}=\frac{y_0-0}{0-x_0}=-\frac{7-x_0}{x_0}$

$x_0y=(x_0-7)(x-x_0)$

$(-3,8)$ is on the line hence $8x_0=(x_0-7)((-3)-x_0)$

$x_0^2+4x_0-21=0$

$(x_0+7)(x_0-3)=0$ since $x_0>0$, $x_0=3$

$y_0=4$ the equation is $3y=-4x+21$

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