Find the equation of the straight lines each passing through the point ( 6 , -2 )and whose sum of the intercepts is 5
Answers
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
Answer:
x-2y=10
2x+3y=6
Step-by-step explanation:
x/a+y/b=1 is the equation.
Put x=6,y=-2,
6/a-2/b=1
a+b=5, or, b=5-a
6/a-2/(5-a)=1
a^2-13a+30=0
a=10,3
So, b=-5,2
Putting values of a and b , we can find the required st.lines