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Question

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a) (-3, -8)

b) $\left( \dfrac{1}{3},\dfrac{-8}{3} \right)$

c) $\left( \dfrac{-10}{3},\dfrac{7}{3} \right)$

d) ( -3, -9)

Answer

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We are given 2 lines in the question and the point where the diagonals intersect. Now we know that in a rhombus opposite sides are parallel and our equations are intersecting so they must belong to adjacent sides.

First, we find the perpendicular distance to both the lines from the given point.

The formula is: - d = $\dfrac{\left| am+bn+c \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$

Where d is the distance and a, b are coefficients of equation of line and (m,n) are the coordinates of the external point

Line 1: - x-y+1=0 and (-1, -2)

d = $\dfrac{\left( -1+2+1 \right)}{\sqrt{2}}$ = $\sqrt{2}$

So, equation of line parallel to this one and other side of the Rhombus is

Distance between parallel lines = $\dfrac{\left| {{c}_{1}}-{{c}_{2}} \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}}}$

$2\sqrt{2}=\dfrac{\left| 1-{{c}_{2}} \right|}{\sqrt{2}}$

$4=\left| 1-{{c}_{2}} \right|$

$c_2$ = 5, -3.

To find out the point we have to check the distance by using both values of c and that gives us

$c_2$ = -3. Equation of parallel line => x-y-3 = 0

Similarly, we find the equation for other line which comes out to be

7x-y+15 = 0

Our final diagram after plotting all the equations and point looks like this:

In the above figure the given lines are drawn and the formed rhombus is shown having vertices A, B, C, D and the point of intersection of the diagonals AC and BD is O which is at (-1,-2). The point D is at the intersection of the lines x-y+1=0 and 7x-y-5=0 and the point B is at the intersection of the lines x-y-3=0 and 7x-y+15=0.

Intersection of the line 7x-y-5 = 0 and x-y-3 = 0 is at the point C whose coordinates are given by $\left( \dfrac{1}{3},-\dfrac{8}{3} \right)$.