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 $