Answer:

GIVEN :–

▪︎ Straight line passing from point (6,-2).

▪︎ Sum of intercepts = 5

TO FIND :–

Equation of straight line .

SOLUTION :–

▪︎ Intercept FORM of line is –

\begin{lgathered}\\ \implies \: { \boxed{ \bold{ \dfrac{x}{a} + \dfrac{y}{b} = 1 }}} \\\end{lgathered}

⟹

a

x

+

b

y

=1

\begin{lgathered}\\ \to \: { \bold{Here \: a \: \: and \: \: b \: \: are \: \: x - intercept \: \: and \: \: y - intercept \: \: respectively.}} \\\end{lgathered}

→Hereaandbarex−interceptandy−interceptrespectively.

▪︎ According to the question line passing from a point (6,-2) , So that –

\begin{lgathered}\\ \implies{ \bold{ \dfrac{6}{a} + \dfrac{( - 2)}{b} = 1 \: \: \: \: - - - - eq.(1)}} \\\end{lgathered}

⟹

a

6

+

b

(−2)

=1−−−−eq.(1)

• And sum of intercepts = 5

\begin{lgathered}\\ \implies{ \bold{ a + b= 5 \: \: \: \: - - - - eq.(2)}} \\\end{lgathered}

⟹a+b=5−−−−eq.(2)

▪︎ Now by eq.(1) and eq.(2) –

\begin{lgathered}\\ \implies{ \bold{ \dfrac{6}{a} + \dfrac{( - 2)}{(5 - a)} = 1 }} \\\end{lgathered}

⟹

a

6

+

(5−a)

(−2)

=1

\begin{lgathered}\\ \implies{ \bold{ \dfrac{6(5 - a) - 2(a)}{a(5 - a)} = 1 }} \\\end{lgathered}

⟹

a(5−a)

6(5−a)−2(a)

=1

\begin{lgathered}\\ \implies{ \bold{ \dfrac{30- 6a- 2a}{a(5 - a)} = 1 }} \\\end{lgathered}

⟹

a(5−a)

30−6a−2a

=1

\begin{lgathered}\\ \implies{ \bold{ \dfrac{30 - 8a}{a(5 - a)} = 1 }} \\\end{lgathered}

⟹

a(5−a)

30−8a

=1

\begin{lgathered}\\ \implies{ \bold{ 30 - 8a = 5a - {a}^{2} }} \\\end{lgathered}

⟹30−8a=5a−a

2

\begin{lgathered}\\ \implies{ \bold{ {a }^{2} - 13A + 30 = 0 }} \\\end{lgathered}

⟹a

2

−13a+30=0

\begin{lgathered}\\ \implies{ \bold{ {a }^{2} - 10a - 3a + 30 = 0 }} \\\end{lgathered}

⟹a

2

−10a−3a+30=0

\begin{lgathered}\\ \implies{ \bold{ a(a - 10) - 3(a - 10 )= 0 }} \\\end{lgathered}

⟹a(a−10)−3(a−10)=0

\begin{lgathered}\\ \implies{ \bold{( a - 3)(a - 10) = 0 }} \\\end{lgathered}

⟹(a−3)(a−10)=0

\begin{lgathered}\\ \implies{ \bold{a = 3 \: , \: a = 10 }} \\\end{lgathered}

⟹a=3,a=10

• Now using eq.(2) –

\begin{lgathered}\\ \implies{ \bold{b = 2 \: , \: b = - 5 }} \\\end{lgathered}

⟹b=2,b=−5

EQUATION :–

(1) When a = 3 , b = 2 :–

\begin{lgathered}\\ \implies \: { \boxed{ \bold{ \dfrac{x}{3} + \dfrac{y}{2} = 1 }}} \\\end{lgathered}

⟹

3

x

+

2

y

=1

(2) When a = 10 , b = -5 :–

\begin{lgathered}\\ \implies \: { \boxed{ \bold{ \dfrac{x}{10} + \dfrac{y}{ (- 5)} = 1 }}} \\\end{lgathered}

⟹

10

x

+

(−5)

y

=1