Q.If the line $\dfrac{x}{a}+\dfrac{y}{b}=1$ passes through the points $(2,-3)$ and $(4,-5)$, then $(a,b)$ is
(A) $(1,1)$
(B) $(-1,1)$
(C) $(1,-1)$
(D) $(-1,-1)$

The key idea is that if a point lies on a line, its coordinates must satisfy the line's equation.

1. Substitute the point $(2,-3)$ into the line equation $\dfrac{x}{a}+\dfrac{y}{b}=1$:

   $\dfrac{2}{a} + \dfrac{-3}{b} = 1 \implies \dfrac{2}{a} - \dfrac{3}{b} = 1 \quad (1)$

2. Substitute the point $(4,-5)$ into the line equation:

   $\dfrac{4}{a} + \dfrac{-5}{b} = 1 \implies \dfrac{4}{a} - \dfrac{5}{b} = 1 \quad (2)$

3. Multiply equation (1) by 2:

   $2 \left( \dfrac{2}{a} - \dfrac{3}{b} \right) = 2(1) \implies \dfrac{4}{a} - \dfrac{6}{b} = 2 \quad (3)$

4. Subtract equation (2) from equation (3):

   $\left( \dfrac{4}{a} - \dfrac{6}{b} \right) - \left( \dfrac{4}{a} - \dfrac{5}{b} \right) = 2 - 1$ …

The problem asks us to find the values of $a$ and $b$ for a line given in intercept form, $\frac{x}{a} + \frac{y}{b} = 1$, which passes through two specific points.

The fundamental concept here is that if a point lies on a line, its coordinates must satisfy the equation of that line. This means that if we substitute the $x$ and $y$ coordinates of a point into the line's equation, the equation must hold true. Since the line passes through two points, we can perform this substitution twice, generating two separate equations. These two equations will form a system that we can then solve to find the unknown values of $a$ and $b$.

Let's apply this concept step-by-step.

1. Formulate equations from the given points.

   The equation of the line is $\frac{x}{a} + \frac{y}{b} = 1$.

   The line passes through $(2, -3)$ and $(4, -5)$.

   • Using the point $(2, -3)$:

     Substitute $x=2$ and $y=-3$ into the line equation:

     $\frac{2}{a} + \frac{-3}{b} = 1$

     This simplifies to:

     $\frac{2}{a} - \frac{3}{b} = 1 \quad \text{(Equation 1)}$

   • Using the point $(4, -5)$:

     Substitute $x=4$ and $y=-5$ into the line equation:

     $\frac{4}{a} + \frac{-5}{b} = 1$

     This simplifies to:

     $\frac{4}{a} - \frac{5}{b} = 1 \quad \text{(Equation 2)}$

2. Solve the system of linear equations.

   We now have a system of two equations with two unknowns, $\frac{1}{a}$ and $\frac{1}{b}$:

   1. $2\left(\frac{1}{a}\right) - 3\left(\frac{1}{b}\right) = 1$
   2. $4\left(\frac{1}{a}\right) - 5\left(\frac{1}{b}\right) = 1$

   To make this system look more familiar, let's substitute $X = \frac{1}{a}$ and $Y = \frac{1}{b}$.

   1. $2X - 3Y = 1$
   2. $4X - 5Y = 1$ …