Question

Find theequation of the straight line passing through (-3,10)
and sum of their intercepts is 8.​

Answer 1

Answer:

Point: ( -3, 10 )

Sum of Intercepts: 8

Let the intercepts be denoted as 'a' ad 'b'

⇒ a + b = 8

According to Point Intercept Form,

⇒ x/a + y/b = 1

⇒ -3/x + 10/b = 1

⇒ -3/a + 10/8-a = 1

Taking LCM we get,

⇒ -3 ( 8 - a ) + 10 a / ( 8 - a ) a = 1  

⇒ 3a - 24 + 10a / 8a - a² = 1

Cross multiplying we get,

⇒ 3a - 24 + 10 = 8a - a ²

⇒ 13a - 24 = 8a - a²

⇒ a² + 13a - 8a - 24 = 0

⇒ a² + 5a - 24 = 0

Solving this equation we get,

⇒ a² + 8a - 3a - 24 = 0

⇒ a ( a + 8 ) -3 ( a + 8 ) = 0

⇒ ( a - 3 ) ( a + 8 ) = 0

⇒ a = 3, -8

If 'a' = 3, then 'b' = 10 - 3 = 7

⇒ Eqn would be: x/3 + y/7 = 1

If 'a' = -8, then 'b' = 10 - ( -8 ) = 18

⇒ Eqn would be: x/-8 + y/18 = 1

Answer 2

AnswEr :

⠀

Let the Intercept be a and b

• Points = ( x,y ) = ( -3,10 )
• Sum of Intercept = (a + b) = 8

» a + b = 8

» b = 8 - a

⠀

• Now A.T.Q. Point Intercept Form

$\Rightarrow \bf{ \dfrac{x}{a} + \dfrac{y}{b} = 1 }$

$\Rightarrow \bf{ \dfrac{ - 3}{a} + \dfrac{10}{(8 - a)} = 1}$

$\Rightarrow \bf{ \dfrac{ - 3(8 - a) + 10a}{a(8 - a)} = 1 }$

$\Rightarrow \bf{ \dfrac{ -24 + 3a + 10a}{8a - {a}^{2} } = 1 }$

• By Cross Multiplication

⇒ - 24 + 3a + 10a = 8a - a²

⇒ 13a - 24 = 8a - a²

⇒ a² + 13a - 8a - 24 = 0

⇒ a² + 5a - 24 = 0

• Splitting Middle Term

⇒ a² + 8a - 3a - 24 = 0

⇒ a(a + 8) - 3(a + 8) = 0

⇒ (a - 3)(a + 8) = 0

⇒ a = 3 ⠀or, ⠀a = - 8

⇒ b = (8 - 3) = 5 ⠀or, ⠀b = {8 + (- 8)} = 16

⠀

Equation Can Be :

⋆ $\large\boxed{\bf{ \dfrac{x}{3} + \dfrac{y}{5} = 1 }}$

⋆ $\large\boxed{\bf{ \dfrac{x}{-8} + \dfrac{y}{16} = 1 }}$