Question

Question 14
The perpendicular bisector of a line segment joining points A(1,5) and B(4,6) cuts Y-axis at
(A) (0, 13)
(B) (0, –13)
(C) (0, 12)
(D) (13, 0)

Answer 1

Firstly, we will plot the points of the line segment on paper and will join them.

We know that, the perpendicular bisector of the line segment AB, bisects the segment AB
i.e., perpendicular bisector of line segment AB, passes through the mid-point of AB.
$\therefore \text{Mid-point of AB}=(\frac{1+4}{2},\frac{5+6}{2})\Rightarrow P=(\frac{5}{2},\frac{11}{2})\left[\begin{array}{c}\because \text{Mid-point of line segment having end - points}\\ (x_1,y_1)and(x_2,y_2)\\ =(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\end{array}\right]$
Now, we will draw a straight line on paper that passes through the mid-point P.
Perpendicular bisector cuts the Y-axis at the point (0,13).
Hence, the required point is (0,13).
Alternate Method
We know that, the equation of line which passes through the points
$(x_1,y_1)and(x_2,y_2)$ will be
$(y-y_1)=\frac{y_2-y_1}{x_2-x_1}(x-x_1)...\left(i\right)Here,x_1=1,y_1=5and{x}_2=4,y_2=6$
So, equation of the line segment joining points A(1,5) and B(4,6) is
$(y-5)=\frac{6-5}{4-1}(x-1)\Rightarrow (y-5)=\frac{1}{3}(x-1)\Rightarrow 3y-15=x-1\Rightarrow 3y=x-14\Rightarrow y=\frac{1}{3}x-\frac{14}{3}...\left(ii\right)\therefore \text{Slope of the segment,}{m}_1=\frac{1}{3}$
If two lines are perpendicular to each other, then the relation between its slopes is
$m_1{m}_2=-1...\left(iii\right)where,m_1=\text{slope of line 1}\text{and}{m}_2\text{= slope of line 2}$
Also, we know that the perpendicular bisector of the line segment is perpendicular on the line
segment.
Let slope of line segment is $m_2$.
$\text{From Eq. (iii),}{m}_1.m_2=\frac{1}{3}.m_2=-1\Rightarrow m_2=-3$
Also, we know that the perpendicular bisector is passing through the mid-point of line segment.
$\therefore \text{Mid-point of line segment}=(\frac{1+4}{2},\frac{5+6}{2})=(\frac{5}{2},\frac{11}{2})\text{Equation of perpendicular bisector,}\text{which has slope (-3) and passes}\text{through the point}(\frac{5}{2},\frac{11}{2})is(y-\frac{11}{2}=(-3\left)(x-\frac{5}{2})\right)$
[Since, equation of line passes through the point $(x_1,y_1)$ and having slope, m $(y-y_1)=m(x-x_1)$
$\Rightarrow (2y-11)=-3(2a-5)\Rightarrow 2y-11=-6+15\Rightarrow 6x+2y=26\Rightarrow 3x+y=13$
If the perpendicular bisector cuts the Y-axis, then put x=0 in Eq. (iv)
$3\times 0+y=13\Rightarrow y=13\text{So, the required point is (0,13).}$