Straight line question

Question

One of the previous iit question If a line passing through (8,2) with negative slope has positive intercepts on co ordinate axes Then find the min value of sum of intercepts? I have the solution and answer is 18.... But if x+y=10 Then sum of intercepts is 20 Why??

Answer 1

Let the line be $y=ax+b$. Then the $y$-intercept is $b$ and $x$-intercept is $-\frac{b}{a}$. So: $$b-\frac{b}{a} \to \text{min}, \ \ \text{subject to} \ 2=8a+b$$ Can you handle the rest? The answer (sum of x and y intercepts) is $12+6=18$. The line is $y=-\frac12x+6$.

Answer 2

we have $$y=mx+n$$ as the searched line, since the Point $(8;2)$ is situated on the line we get $$y=m(x-2)+2$$ the Points of intercepts are $$A(0;-8m+2)$$ $$B(8-\frac{2}{m};0)$$ so the searched sum is given by $$f(m)=-8m+2+8-\frac{2}{m}$$ Can you finish?

Answer 3

The equation of the line can be written $y = -kx + c$ where $k>0$. Given that $(8,2)$ lies on the line, $2 = -8k + c \implies c = 8k + 2$, so the equation can now be rewritten $y = -kx + (8k+2)$.

Set $x$ and then $y$ to zero in turn to solve for the intercepts:

The $y$ intercept $y_0 = 8k+2$.

The $x$ intercept $x_0 = 8 + \frac 2k$.

The sum $S(k) = x_0 + y_0 = 10 + 8k + \frac 2k$.

$S'(k) = 8 - \frac{2}{k^2}$. Setting $S'(k) = 0$ gives $k = \frac 12$ (ignore the negative root since we originally stipulated that $k>0$ for a negative slope).

Since $S''(\frac 12) > 0$, this stationary value is a minimum, which can be computed to be $S(\frac 12) = 10 + 4 + 4 = 18$.