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!
Best Answer
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 $x_0$.