Question

find the equation of a straight line passing through the point (6,-2) and whose sum of positive intercept is 5​

Answer 1

Given :

The given point through which the straight line passes is :

(6 , -2)

Sum of positive intercepts of this straight line = 5

To Find :

What is the equation of this line ?

Solution :

Since the equation of a line in slope intercept form is given as :

$\frac{x}{a } + \frac{y}{b} = 1$   - (1)

Here a and b are the intercepts cut by the line on the x and y axis respectively .

Also we have :

a + b = 5

Or, b = (5-a)

Now from equation :

$\frac{x}{a} + \frac{y}{5-a}$ = 1

Now since this line passes through (6 , -2 ) so this point must satisfy the equation of this line , so :

$\frac{6}{a} - \frac{2}{5-a} = 1$

Or, $\frac{6(5-a)-2a}{a(5-a)}=1$

Or ,$30 - 6a - 2a = 5a - a^2$

Or , $a^2 - 13a + 30 = 0$

So , $a = \frac{13 \pm \sqrt{13^2 - 4 \times 1 \times 30}}{2\times 1}$

    $a = \frac{13 \pm \sqrt{49}}{2}$

    a =$\frac{13 \pm7}{2}$

So , a = 10 or a = 3

Hence , b =-5 or b = 2

Since it is given that the intercepts are positive , so :

a = 3 and b = 2

So , equation of the line is :

$\frac{x}{3}+ \frac{y}{2} = 1$

Hence the equation of the required line is $\frac{x}{3}+ \frac{y}{2} = 1$