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Question: The point \[(2k – 8,\ 10)\] lies on y axis . find the value of \[k\] ?...

The point (2k8, 10)(2k – 8,\ 10) lies on y axis . find the value of kk ?

Explanation

Solution

In this question, we need to find out the value of kk in the point (2k8, 10)(2k – 8,\ 10) and also given that the point lies on y axis. Here we will use the concepts of the Cartesian plane that if the point lies on the y axis then the x value of the given point must be zero. First we can consider any point on the y axis to be (0, n)(0,\ n) where nn is any integer. Then on equating the point and simplifying , we can get the value of kk.

Complete step by step solution:
Given the point (2k8, 10)(2k – 8,\ 10) lies on y axis.
Here we need to find the value of kk.
Given that the point lies on y axis. We need to know that if the point lies on the y-axis then the x value of the given point must be zero.
Let us consider any point on the y axis be (0, n)(0,\ n) where nn is any integer.
Since the point on the y axis is (0, n)(0,\ n) , it follows as
(0, n) = (2k8, 10)(0,\ n)\ = \ (2k – 8,\ 10)
On equating both the points,
We get,
 0=2k8\Rightarrow \ 0 = 2k – 8 and n=10n = 10
Now we need to find the value of kk,
 0=2k8\Rightarrow \ 0 = 2k – 8
On rewriting the terms,
We get,
2k=8\Rightarrow 2k = 8
On dividing both sides by 22 ,
We get,
 k=82\Rightarrow \ k = \dfrac{8}{2}
On simplifying,
We get,
 k=4\Rightarrow \ k = 4
Thus we get the value of kk as 44 .
The value of kk as 44 .

Note:
In order to solve these types of questions , we must know the orientation of all the axes. The horizontal axis is often called the x axis and the vertical axis is called the y axis. The upper region is known as the positive y axis whereas the lower region is known as the positive y axis. The leftmost part is known as the positive x axis whereas the rightmost part is known as the positive x axis. It is important to remember that if x coordinate of a point is zero then that point lies on the y axis and if y coordinate of a point is zero then that point lies on the x axis.