Question

Find the value of k for which the point $\left( {1, - 2} \right)$ lies on the graph of the linear equation $x - 2y + k = 0$.

Answer 1

Hint: To find the value of k in equation $x - 2y + k = 0$ at point $\left( {1, - 2} \right)$, we need to substitute the value of x and y as 1 and -2 in the given equation. After substituting these values, simplify the obtained equation and you will get the value of k.

Complete step by step solution:
In this question, we are given a linear equation with variable x and y and constant k and we need to find the value of constant k at point (1, -2).
Given equation is: $x - 2y + k = 0$ - - - - - - - - - - - - - (1)
So, to find the value of k in equation (1), we need to put the values of x and y as 1 and -2 in equation (1).
So, therefore, on substituting the values of x and y in equation (1), we get
$
   \Rightarrow \left( 1 \right) - 2\left( { - 2} \right) + k = 0 \\
   \Rightarrow 1 + 4 + k = 0 \\
 $
$ \Rightarrow k + 5 = 0$ - - - - - - - - - - - - - - - - (2)
Now, subtracting 5 on both LHS and RHS of equation (2), we get
  $
   \Rightarrow k + 5 - 5 = 0 - 5 \\
   \Rightarrow k = - 5 \\
 $
Hence, the value of k for equation $x - 2y + k = 0$ at point $\left( {1, - 2} \right)$ is $-5$.

Note:
We can cross check our answer by putting the value of $k$ as $-5$ in the given equation and then substituting the values of $x$ and $y$ as $1$ and $-2$. If we get our answer as 0 then our answer is correct and if not then our answer is incorrect.