Question

find the equation of the straight line which make intercept whose sum is 5 & product is 6 ?

Answer 1

Step-by-step explanation:

the required equation of the straight line is x/a+y/b=1

given ab=6

b=6/a_______(1)

given a+b=5

a+6/a=5

a^2+6=5a

a^2-5a+6=0

a^2-3a-2a+6=0

a(a-3)-2(a-3)=0

a-2=0. ,a-3=0

a=2, a=3

at a=2

substitute a=2 in equation 1

b=6/a

b=6/2

b=3

equation of the line is x/a+y/b=1

x/2+y/3=1

3x+2y=6

at x=3

substitute a=3in equation 1

b=6/a

b=6/3

b=2

equation of the line is x/a+y/b=1

x/3+y/2=1

2x+3y=6

Answer 2

Answer:

Step-by-step explanation:

Concept:

A straight line's general equation is $y = mx + c$, where m is the gradient and $y = c$ is the value at which the line intersects the y-axis. This value c is known as the y-axis intercept.

Given:

The sum is $5$ and product is $6$.

Find:

find the straight line equation that makes intercept

Solution:

Consider that $\mathrm{a}$ and $\mathrm{b}$ be $\mathrm{x}$ - intercepts and $\mathrm{y}$-intercepts

The straight line equation is

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

We know that

Sum of intercepts $(a+b)=5$

Product of the intercepts $(a b)=6$

$\begin{aligned}&(a-b)^{2}=(a+b)^{2}-4 a b \\&=5^{2}-4(6) \\&=25-24 \\&=1\end{aligned}$

Simplify

$a+b=5..........(1)\\a-b=1.............(2)\\a-b=-1.............(3)$

Now lets solve equation 1 and 2.

$a+b=5\\a-b=1$

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$2a=6\\a=\frac{6}{2} \\a=3$

Insert value of 'a' in equation (1)

$\begin{aligned}&3+b=5 \\&3+b=5 \\&b=5-3 \\&b=2\end{aligned}$

Now, Solve both the equations (1) and (2)

$a+b=5 \\a-b=-1$

-------------------

$2 a=4 \\a=\frac{4}{2} \\a=2$

Substitute the value of (a) in (1)

$2+b=5 \\a=\frac{4}{2} \\a=2$

Insert value of  a in (1)

$\begin{aligned}&2+b=5 \\&b=5-2 \\&b=3\end{aligned}$

Hence, Equation of the required line is

$\begin{gathered}\frac{x}{3}+\frac{y}{2}=1 \rightarrow 2 x+3 y=6 \\(\text { or }) \\\frac{x}{2}+\frac{y}{3}=1 \rightarrow 3 x+2 y=6\end{gathered}$

#SPJ3