At what point does the graph of the linear equation x + y = 5 meet a line which is parallel to the
y-axis, at a distance 2 units from the origin and in the positive direction of x-axis.
Answers
Solution:
At point (2,3) the graph of the linear equation x + y = 5 meet a line which is parallel to the y-axis, at a distance 2 units from the origin and in the positive direction of x-axis.
Explanation:
The given line is
Let us write this equation in intercept form. In order to write in this form, we can divide both sides of the equation by 5.
Thus, the x-intercept is 5 and the y-intercept is 5.
Now, any line parallel to the y axis is in the form x=k.
Since, this line is at a distance of 2 units from the origin hence and in the positive x-axis.
Hence, the equation of the line is x=2.
The intersection point of the line x=2 and the given line will give us the required point.
Plugging x=2 in the given equation, we get
Therefore, the required point is (2,3).
We can see this in the attached file as well.
Answer:
Step-by-step explanation:
At point (2,3) the graph of the linear equation x + y = 5 meet a line which is parallel to the y-axis, at a distance 2 units from the origin and in the positive direction of x-axis.
Explanation:
The given line is x+y=5x+y=5
Let us write this equation in intercept form. In order to write in this form, we can divide both sides of the equation by 5.
\frac{x}{5}+\frac{y}{5}=1
5
x
+
5
y
=1
Thus, the x-intercept is 5 and the y-intercept is 5.
Now, any line parallel to the y axis is in the form x=k.
Since, this line is at a distance of 2 units from the origin hence and in the positive x-axis.
Hence, the equation of the line is x=2.
The intersection point of the line x=2 and the given line will give us the required point.
Plugging x=2 in the given equation, we get
\begin{gathered}2+y=5\\\\y=3\end{gathered}
2+y=5
y=3
Therefore, the required point is (2,3).
We can see this in the attached file as well.
