Question

Find the equation of the straight lines each passing through the point (6,-2) and whose sum of the intercept is 5.

Answer 1

Let the equation of straight line be (x/a) +(y/b) = 1 -------- (i)
a + b = 5 (given in the question) ------ (ii)
Putting the value of $\times$ and& y coordinates in (i) we get
(6/a) + (-2/b) = 1 ---------- (iii)
Substituting the value of b in (iii), we get
6(a-5) - 2a = a(a-5)
$\Rightarrow$ 6a – 30 -2a = $a^2$ – 5a
$\Rightarrow$ $a^2$ – a +30 = 0
$\Rightarrow$ $a^2$ -6a +5a +30 = 0
$\Rightarrow$a(a-6) + 5(a-6)= 0
$\Rightarrow$ (a-6)(a+5) = 0
$\Rightarrow$ a = 6, a= -5
Putting the value of a in equation (i), we get
When a is 6
$\Rightarrow$ 6 + b = 5, b = -1
When a is -5
$\Rightarrow$ -5 + b = 5,
$\Rightarrow$ b = 10.

So, the required equations are

$\frac{x}{6}+\frac{y}{-1}=1$ where a and b are 6 and -1 respectively.
$\frac{x}{-5}+\frac{y}{10}=1$ where a and b are -5 and 10 respectively.