Question

Find the equations of the straight lines each passing through the point $\left( 6,-2 \right)$ and whose sum of the intercept is 5.

Answer 1

Hint: First, before proceeding for this, we must know the following general equation of the line with x-intercept as a and y-intercept as b is given by$\dfrac{x}{a}+\dfrac{y}{b}=1$. Then, we have the condition given in the question as the sum of the intercepts a and b is 5. Then, we are given that the line is passing through the point (6,-2) to get the value of a and b and hence get the final equation of line.

Complete step-by-step answer:
In this question, we are supposed to find the equations of the straight lines each passing through the point $\left( 6,-2 \right)$ and whose sum of the intercept is 5.
So, before proceeding for this, we must know the following general equation of the line with x-intercept as a and y-intercept as b is given by:
$\dfrac{x}{a}+\dfrac{y}{b}=1$
Then, we have the condition given in the question as the sum of the intercepts a and b is 5.
$a+b=5$
Now by using the above equation to get the value of b as:
$b=5-a$
Now, by substituting the value of b in the general equation of the line, we get:
$\begin{align}
  & \dfrac{x}{a}+\dfrac{y}{5-a}=1 \\
 & \Rightarrow \dfrac{x\left( 5-a \right)+ya}{a\left( 5-a \right)}=1 \\
 & \Rightarrow x\left( 5-a \right)+ay=a\left( 5-a \right) \\
\end{align}$
Now, we are given that the line is passing through the point (6,-2).
So, by substituting the value of x as 6 and y as -2, we get:
$6\left( 5-a \right)-2a=a\left( 5-a \right)$
Now, by solving the above equation to get the value of a as:
$\begin{align}
  & 30-6a-2a=5a-{{a}^{2}} \\
 & \Rightarrow {{a}^{2}}-13a+30=0 \\
\end{align}$
Now, by using the quadratic formula to get the value of a for general equation $a{{x}^{2}}+bx+c=0$ as:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
So, the value of a is as:
$\begin{align}
  & a=\dfrac{13\pm \sqrt{{{13}^{2}}-4\times 1\times 30}}{2} \\
 & \Rightarrow a=\dfrac{13\pm \sqrt{169-120}}{2} \\
 & \Rightarrow a=\dfrac{13\pm \sqrt{49}}{2} \\
 & \Rightarrow a=\dfrac{13\pm 7}{2} \\
 & \Rightarrow a=\dfrac{13+7}{2},\dfrac{13-7}{2} \\
 & \Rightarrow a=10,3 \\
\end{align}$
So, we get two values of a as 3 and 10.
Now to get the corresponding values of the b are:
For $a=3, 3+b=5$ gives $b=2$
For $a=10, 10+b=5$ gives $b=-5$
Now, substituting the values of a as 3 and b as 2, we get the equation as:
$\dfrac{x}{3}+\dfrac{y}{2}=1$
Similarly, the equation after substituting the value of a as 10 and b as -5:
$\dfrac{x}{10}-\dfrac{y}{5}=1$
Hence, the equation of the line passing through point (6,-2) and whose sum of intercept is 5 is given by $\dfrac{x}{3}+\dfrac{y}{2}=1$ and $\dfrac{x}{10}-\dfrac{y}{5}=1$.

Note: Now, to solve these types of questions we need to know some of the basic representations of the straight line. So, some of the basic representations of the straight line are:
$\dfrac{x}{a}+\dfrac{y}{b}=1$ where a is x-intercept and b is y-intercept.
$y=mx+c$ where m is slope and c is y-intercept of straight line.