Question

Points $A\left( {4,1} \right)$ lies on a line :
$\left( A \right){\text{ }}x + 2y = 5$
$\left( B \right){\text{ }}x + 2y = 6$
$\left( C \right){\text{ }}x + 2y = 16$
$\left( D \right){\text{ }}x + 2y = - 6$

Answer 1

Hint : If a point $\left( {x,y} \right)$ lies on a line, then the equation of the line must be satisfied by the given point $\left( {x,y} \right),$ means by substituting the values of $x{\text{ and }}y$ , the equation should hold true. Hence to solve this question we will put the given values of $x{\text{ and }}y$ in all the given equations and the equation which will satisfy these values will be our answer. The equation of a line in slope intercept form is written as, $y = mx + c$ ; where $m$ is the slope or gradient of the line and $c$ is called the intercept on the $y - axis$ . The slope or gradient of a line is calculated by the formula , $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$ where $\left( {{x_1},{y_1}} \right){\text{ and }}\left( {{x_2},{y_2}} \right)$ are called the starting and end points of the line respectively.

Complete step-by-step answer :
The given point is : $\left( {x,y} \right) = \left( {4,1} \right)$
The quadrant system is shown in the figure below:

According to the quadrant system shown above, the given point $A\left( {4,1} \right)$ lies in the first quadrant as both the values are positive.
Now, we will have to check for all the equations given in the question one by one:
$\left( A \right){\text{ }}x + 2y = 5$
$\Rightarrow x = 4{\text{ and }}y = 1$ (Given)
Put the values of $x{\text{ and }}y$ in the given equation, we get;
${\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \Rightarrow {\text{ 4}} + \left( {2 \times 1} \right) = 6$
${\text{R}}{\text{.H}}{\text{.S}}{\text{. = 5}}$
$\Rightarrow 6 \ne 5$
Hence, ${\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.S}}{\text{.}}$
Therefore, option A can not be the correct answer.

$\left( B \right){\text{ }}x + 2y = 6$
$\Rightarrow x = 4{\text{ and }}y = 1$ (Given)
Put the values of $x{\text{ and }}y$ in the given equation, we get;
${\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \Rightarrow {\text{ 4}} + \left( {2 \times 1} \right) = 6$
R.H.S. = 6
$\Rightarrow 6 = 6$
Hence, ${\text{L}}{\text{.H}}{\text{.S}}{\text{. = R}}{\text{.H}}{\text{.S}}{\text{.}}$
Therefore, option B is the correct answer for this question.

$\left( C \right){\text{ }}x + 2y = 16$
$\Rightarrow x = 4{\text{ and }}y = 1$ (Given)
Put the values of $x{\text{ and }}y$ in the given equation, we get;
${\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \Rightarrow {\text{ 4}} + \left( {2 \times 1} \right) = 6$
${\text{R}}{\text{.H}}{\text{.S}}{\text{. = 16}}$
$\Rightarrow 6 \ne 16$
Hence, ${\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \ne {\text{ R}}{\text{.H}}{\text{.S}}{\text{.}}$
Therefore, option C can not be the correct answer.

$\left( D \right){\text{ }}x + 2y = - 6$
$\Rightarrow x = 4{\text{ and }}y = 1$ (Given)
Put the values of $x{\text{ and }}y$ in the given equation, we get;
${\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \Rightarrow {\text{ 4}} + \left( {2 \times 1} \right) = 6$
${\text{R}}{\text{.H}}{\text{.S}}{\text{. = }} - 6$
$\Rightarrow 6 \ne - 6$
Hence, ${\text{L}}{\text{.H}}{\text{.S}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.S}}{\text{.}}$
Hence point $A\left( {4,1} \right)$ lies on a line whose equation is given by: $x + 2y = 6$
So, the correct answer is “Option B”.

Note : We have seen the slope intercept form of a straight line with a gradient and intercept on the $y - axis$ above. But in this question, we are given the equation of a straight line in a different form. For example: $x + 2y = - 6$ , then how can we represent this equation in the standard form. By rearranging the equation as $2y = - x - 6$ or $y = - \dfrac{1}{2}x - 3$ . Now comparing this with the standard equation i.e. $y = mx + c$ , we can say that $m = - \dfrac{1}{2}$ and $y - {\text{intercept or }}c = - 3$