Question

18. Find the equation of the line passing through (22, -6) and having intercept on x-axis
exceeds the intercept on y-axis by 5.

Answer 1

Final Answer :
$\frac{x}{10} + \frac{y}{5} = 1 \\ \frac{x}{11} + \frac{y}{6} = 1$

Steps:
1) We have,
Point on line P(x,y) = (22,-6)
Let the x-intercept and y-intercept be 'a'and 'b' respectively .
Then,
According to the question,
a = b + 5 --------(2)

2) We know,
Intercept Form of line,
$\frac{x}{a} + \frac{y}{b} = 1$
Using equation (1) and (2) which satisfies the equation.
$\frac{22}{b + 5} - \frac{6}{b} = 1 \\ = > 22 \times b - 6 \times (b + 5) = b \times (b + 5) \\ = > 22 \times b - 6b - 30 = {b}^{2} + 5b \\ = > {b}^{2} - 11b + 30 = 0 \\ = > (b - 5)(b - 6) = 0 \\ = > b = 5 \: \: or \: \: b \: = 6$

3) Then,
For b = 5 , a = b+ 5 = 5+5 =10
For b = 6 , a = b + 5 = 6 + 5 = 11

4) We have two possibilities,
=>
a = 10, b = 5
Equation of line: see pic 1
$\frac{x}{10} + \frac{y}{5} = 1$
Equation of Line : see pic 2
=> a = 11 , b = 6 :
$\frac{x}{11} + \frac{y}{6} = 1$

Answer 2

Solution:-

given by:- the equation of the line passing through (22, -6) and having intercept on x-axis
exceeds the intercept on y-axis by 5.

we have:-

Let x -intercept be= a

y-intercept be =b

accourding to quetion:-

》x -intercept (a = b + 5)

Intercept form:-

》(x/a) + (y/b) = 1

》(22/(b+5)) + (-6/b) = 1

》[22 b - 6(b+5)]/b(b+5) = 1

》(22 b - 6 b - 30)/b^2 + 5b = 1

》(16 b - 30)/b^2 + 5b = 1

》16 b - 30 = b^2 + 5b

》b^2 + 5 b - 16 b  - 30 = 0

》b^2 - 11 b  - 30 = 0

》(b - 6) (b - 5) = 0

》b = 6 and b = 5

》a = 6 + 5   ,      a = 5 + 5

》a = 11      ,       a = 10

》a = 11 , b = 6

》(x/11) + (y/6) = 1

》6 x + 11 y = 66

》6 x + 11  y - 66 = 0

》a = 10 , b = 5

》(x/10) + (y/5) = 1

》 x + 2 y = 10

》x + 2 y - 10 = 0

The required equations are
(x + 2 y - 10 = 0 )or (6 x + 11  y - 66 = 0) ans

☆i hope its help☆