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Question: The points \[A\left( {7,3} \right)\] and \[C\left( {0, - 4} \right)\] are two opposite vertices of a...

The points A(7,3)A\left( {7,3} \right) and C(0,4)C\left( {0, - 4} \right) are two opposite vertices of a rhombus ABCD, find the equation of the diagonal BD.

Explanation

Solution

Consider a point MM as the midpoint of diagonals ACAC and BDBD, and find its coordinates using the midpoint line segment formula to get the required answer. Diagonals in the rhombus are perpendicular to each other. Use slope point form to get the required line equation of the diagonal BD. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer :
Let MM be the midpoint of diagonals ACAC and BDBD as shown in the below figure:

Given points are A(7,3)A\left( {7,3} \right) and C(0,4)C\left( {0, - 4} \right)
We know that the midpoint of the points (x1,y1) and (x2,y2)\left( {{x_1},{y_1}} \right){\text{ and }}\left( {{x_2},{y_2}} \right) is given by (x1+x22,y1+y22)\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right).
Therefore, coordinates of point MM are:
M(7+02,342)=(72,12)M\left( {\dfrac{{7 + 0}}{2},\dfrac{{3 - 4}}{2}} \right) = \left( {\dfrac{7}{2},\dfrac{{ - 1}}{2}} \right)
As we know that in a rhombus diagonal are perpendicular to each other i.e., BDACBD \bot AC
So, the slopes of BD and AC are also perpendicular.
The slope of the line joining points (x1,y1) and (x2,y2)\left( {{x_1},{y_1}} \right){\text{ and }}\left( {{x_2},{y_2}} \right) is given by y2y1x2x1\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}
So, slope of the line AC =4307=77=1 = \dfrac{{ - 4 - 3}}{{0 - 7}} = \dfrac{{ - 7}}{{ - 7}} = 1
Let mm be the slope of the line BD.
We know that the condition of perpendicularity of the two slopes m1{m_1} and m2{m_2} is m1×m2=1{m_1} \times {m_2} = - 1.
As the slopes of BD and AC are perpendicular, we have

m×1=1 m=1  \Rightarrow m \times 1 = - 1 \\\ \therefore m = - 1 \\\

We know that the line equation of a line with slope mmand is passing through a point (x1,y1)\left( {{x_1},{y_1}} \right) is given by yy1=m(xx1)y - {y_1} = m\left( {x - {x_1}} \right). This formula is called a slope point form.
Since the point M(72,12)M\left( {\dfrac{7}{2},\dfrac{{ - 1}}{2}} \right) is passing through the line BD with slope 1 - 1, we have the line equation as

y(12)=1(x72) y+12=x+72 x+y=7212=62=3 x+y=3  \Rightarrow y - \left( {\dfrac{{ - 1}}{2}} \right) = - 1\left( {x - \dfrac{7}{2}} \right) \\\ \Rightarrow y + \dfrac{1}{2} = - x + \dfrac{7}{2} \\\ \Rightarrow x + y = \dfrac{7}{2} - \dfrac{1}{2} = \dfrac{6}{2} = 3 \\\ \therefore x + y = 3 \\\

Note : The midpoint of the points (x1,y1) and (x2,y2)\left( {{x_1},{y_1}} \right){\text{ and }}\left( {{x_2},{y_2}} \right) is given by (x1+x22,y1+y22)\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right). The line equation of a line with slope mmand is passing through a point (x1,y1)\left( {{x_1},{y_1}} \right) is given by yy1=m(xx1)y - {y_1} = m\left( {x - {x_1}} \right). This formula is called a slope point form.