Question

equation of straight line passing through (1,-1) and having the difference of x and y intercept as 4 is​

Answer 1

Answer:

x intercept a

yintercept b

given a-b=4

equation of st.line x/a+y/b=1

equation passes through the pt (1,-1)

therefore 1/a-1/a-4=1

a-4-a=a^2-4a

a^2-4a+4=0

Answer 2

Given,

A straight line passing through (1, -1),

Difference of x- and y-intercept = 4

To find,

Equation of the line.

Solution,

The equation of a line in intercept form is given by,

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

Where 'a' and 'b' are x- and y-intercepts respectively.

It is given here that difference between the x- and the y-intercept is 4. So,

a - b = 4

⇒ b = a - 4.

We can substitute this in the above equation, thus,

$\frac{x}{a} +\frac{y}{a-4} =1$

To find the value of intercept 'a' substitute the point through which it passes, that is x = 1, and y = -1. So,

$\frac{1}{a} +\frac{-1}{a-4} =1$

⇒ $\frac{1}{a} -\frac{1}{a-4} =1$

⇒ $\frac{a-4-a}{a(a-4)} =1$

Rearranging and simplifying,

$a^2-4a =-4$

⇒ $a^2-4a+4=0$

So we get a quadratic equation in 'a'. Solving this equation, we get,

a = 2, 2. That is both the roots are equal.

So, the x-intercept is

a = 2,

now, y-intercept will be

b = a - 4

= 2 - 4

b = -2.

Putting these intercept values in the equation for intercept form,

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

Therefore, the equation of the straight line will be $\frac{x}{2} +\frac{y}{-2} =1.$