Question

From the point $A\left( { - 5,3} \right)$ , a segment is drawn to meet the Y-axis in B. Find the equation of the locus of the midpoint of segment AB.

Answer 1

Hint:According to the given information assume the coordinates of point B with x-coordinate as zero. Now find the coordinates of the midpoint of the line joining points A and B using section formula. Notice that the coordinate of the x-axis is a constant value. Use it to define the locus.

Complete step-by-step answer:
According to the given information in the question, we have a line segment AB with point A at $\left( { - 5,3} \right)$ and point B on the y-axis. With this information, we need to find the locus of the midpoint of line segment AB.
In geometry, a locus is a set of all points (commonly, a line, a line segment, a curve or a surface), whose location satisfies or is determined by one or more specified conditions. And in this case, the condition is that the point should be the midpoint of line segment AB with a varying position of point B on the y-axis.
Since the position of point B is not fixed but it lies on the y-axis, we can assume that the coordinates of the point B as $\left( {0,y} \right)$ .
Now using section formula, according to which the coordinates of the point which divides the line segment joining the points $A\left( {{x_1},{y_1}} \right)$ and $B\left( {{x_2},{y_2}} \right)$ , internally, in the ratio ${m_1}:{m_2}$ are:
$ \Rightarrow \left( {\dfrac{{{m_1}{x_2} + {m_2}{x_1}}}{{{m_2} + {m_1}}},\dfrac{{{m_1}{y_2} + {m_2}{y_1}}}{{{m_2} + {m_1}}}} \right)$
The midpoint of the line segment divides the line segment in the ratio $1:1$ . From section formula, we can write that:
\[ \Rightarrow \left( {\dfrac{{1 \times {x_2} + 1 \times {x_1}}}{{1 + 1}},\dfrac{{1 \times {y_2} + 1 \times {y_1}}}{{1 + 1}}} \right) = \left( {\dfrac{{{x_2} + {x_1}}}{2},\dfrac{{{y_2} + {y_1}}}{2}} \right)\]
So, the coordinates of the midpoint of AB will be given by $\left( {\dfrac{{ - 5 + 0}}{2},\dfrac{{3 + y}}{2}} \right) = \left( {\dfrac{{ - 5}}{2},\dfrac{{3 + y}}{2}} \right)$
Therefore, the general coordinates of the required point are $\left( {\dfrac{{ - 5}}{2},\dfrac{{3 + y}}{2}} \right)$
As we know that the point $B\left( {0,y} \right)$ can lie anywhere on the whole y-axis, this implies \[y \in R \Rightarrow \dfrac{{3 + y}}{2} \in R\]
The coordinates of the required point can take all the values for y-axis but the x-coordinate will always be constant, i.e. $\dfrac{{ - 5}}{2}$ .
$ \Rightarrow x = \dfrac{{ - 5}}{2}$ , which represents a straight vertical line parallel to the y-axis
Hence, the locus of the midpoint of segment AB will be $x = \dfrac{{ - 5}}{2}$

Note:Notice here we checked for the domain of $y$ because we need to prove that the y-coordinate of the midpoint of AB can attain all the values in the y-axis. This eventually helped us to prove that the locus of the required point is a straight line. Be careful while solving the section formula for midpoint. Here the order of both coordinates matter.