Question

(-2,-1) and (4,-5) are the co-ordinates of vertices B and D respectively of rhombus ABCD. Find the equation of the diagonal AC.

Answer 1

$\textsf{\large{\underline{Solution}:}}$

Given That:

$\rm: \longmapsto Coordinates \: Of \: B = ( - 2 ,- 1)$

$\rm: \longmapsto Coordinates \: Of \: D= (4 ,- 5)$

Therefore:

$\rm: \longmapsto Slope \: of \: BD \: (m_{1}) = \dfrac{ \Delta y}{ \Delta x}$

$\rm: \longmapsto Slope \: of \: BD \: (m_{1})= \dfrac{ - 5 + 1}{4 + 2}$

$\rm: \longmapsto m_{1}= \dfrac{ - 4}{6}$

$\rm: \longmapsto m_{1}= \dfrac{ - 2}{3}$

We know that: Diagonals of a rhombus bisect each other at right angle.

Therefore, the product of the slope of AC and BD is -1.

Let slope of AC be m.

Therefore:

$\rm: \longmapsto mm_{1}= - 1$

$\rm: \longmapsto m \times \dfrac{ - 2}{3} = - 1$

$\rm: \longmapsto m =\dfrac{3}{2}$

Now, O is the midpoint of BD. Therefore:

$\rm: \longmapsto Coordinates \: Of \: O= \bigg( \dfrac{ - 2 + 4}{2} , \dfrac{ - 1 - 5}{2} \bigg )$

$\rm: \longmapsto Coordinates \: Of \: O= (1 , - 3 )$

Therefore, equation of line passing through (1, -3) with 3/2 as slope will be:

$\rm: \longmapsto (y - y_{1}) = m(x - x_{1})$

$\rm: \longmapsto (y + 3) = \dfrac{3}{2} (x - 1)$

$\rm: \longmapsto 2(y + 3) = 3(x - 1)$

$\rm: \longmapsto 2y +6= 3x - 3$

$\rm: \longmapsto 2y = 3x - 9$

$\rm: \longmapsto 3x - 2y - 9 = 0$

★ Which is our required answer.

$\textsf{\large{\underline{More To Know}:}}$

1. Slope Intercept form.

$\rm:\longmapsto y=mx+c$

2. Point slope form.

$\rm:\longmapsto y-y_{1}=m(x-x_{1})$

3. Two point form.

$\rm:\longmapsto y-y_{1}=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})$