Question

2. The point A has coordinates (-1, -5) and the point B has coordinates (7, 1). The perpendicular bisector of AB meets the x-axis at C and the y-axis at D. Calculate the length of CD. ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

The point A has coordinates (-1, -5) and the point B has coordinates (7, 1).

We know,

The slope of line segment joining the points (a, b) and (c, d) is represented by m and given by

$\rm :\longmapsto\:\boxed{ \tt{ \: \: m \: = \: \frac{d - b}{c - a} \: \: }}$

So, Slope of AB is given by

$\rm :\longmapsto\:Slope \: of \: AB = \dfrac{1 - ( - 5)}{7 - ( - 1)} = \dfrac{6}{8} = \dfrac{3}{4}$

Also, we know,

Midpoint of line segment joining the points (a, b) and (c, d) is given by

$\rm :\longmapsto\:\boxed{ \tt{ \: (x,y) = \bigg(\dfrac{a + c}{2} , \: \dfrac{b + d}{2} \bigg) \: }}$

So, Let we assume that C be the midpoint of line segment joining the point A having coordinates (-1, -5) and the point B having coordinates (7, 1).

So, Coordinates of C (x, y) is

$\rm :\longmapsto \: (x,y) = \bigg(\dfrac{ - 1 + 7}{2} , \: \dfrac{ - 5 + 1}{2} \bigg) \:$

$\rm :\longmapsto \: (x,y) = \bigg(\dfrac{6}{2} , \: \dfrac{ - 4}{2} \bigg) \:$

$\rm :\longmapsto \: (x,y) = (3, \: - 2)$

So, Coordinates of C = (3, - 2).

Now,

Let we assume that line l be the perpendicular bisector of AB.

So, Slope of line l is given by

$\rm :\longmapsto\:Slope \: of \: l = - \dfrac{1}{Slope \: of \: AB}$

$\bf\implies \:\:Slope \: of \: l = - \dfrac{4}{3}$

We know, Equation of line passing through the point (a, b) having slope m is given by

$\rm :\longmapsto\:\boxed{ \tt{ \: y - b = m(x - a) \: }}$

So, Equation of line l passing through the point (3, - 2) and having slope - 4/3 is given by

$\rm :\longmapsto\:y - ( - 2) = - \dfrac{4}{3}(x - 3)$

$\rm :\longmapsto\:y + 2= - \dfrac{4}{3}(x - 3)$

$\rm :\longmapsto\:3y + 6= - 4x + 12$

$\rm :\longmapsto\:4x + 3y = 12 - 6$

$\red{\bf :\longmapsto\:4x + 3y = 6}$

Now, this line meet y - axis

Substituting 'x = 0' in the given equation, we get

$\red{\bf :\longmapsto\:4(0) + 3y = 6}$

$\red{\bf :\longmapsto\:0 + 3y = 6}$

$\red{\bf :\longmapsto \: 3y = 6}$

$\red{\bf :\longmapsto \: y = 2}$

Now, this line meet x - axis

Substituting 'y = 0' in the given equation, we get

$\red{\bf :\longmapsto\:4x + 3(0) = 6}$

$\red{\bf :\longmapsto\:4x + 0 = 6}$

$\red{\bf :\longmapsto\:4x = 6}$

$\red{\bf :\longmapsto\:x = 1.5}$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf \: 2\\ \\ \sf 1.5 & \sf 0 \end{array}} \\ \end{gathered}$