A line through the midpoint of the sides $AB$ and $AD$ of a rhombus $ABCD,$whose one diagonal is $3x-4y+5=0$ and one vertex is $A(3,1)$ is

Question

The equation of a line through the midpoint of the sides $AB$ and $AD$ of a rhombus $ABCD,$whose one diagonal is $3x-4y+5=0$ and one vertex is $A(3,1)$ is $ax+by+c=0$.Find the absolute value of $a+b+c$,where $a,b,c$ are integers expressed in lowest form.

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My try:
As $A(3,1)$ does not satisfy $3x-4y+5=0$,so the diagonal $BD$ has equation $3x-4y+5=0$.As $AC$ and $BD$ are perpendicular,so equation of $AC$,i found is $4x+3y-15=0$,then i found point of intersection of the two diagonals is $O(\frac{9}{5},\frac{13}{5})$.I want to find the coordinates of $B$ and $D$,so that i can find out the equation of the required line.But for that i need $BO$.

I am stuck here.Please help me.Thanks.

Answer 1

As you wrote, the equation of the line $BD$ is $3x-4y+5=0$, and the intersection point of the two diagonals is $(9/5,13/5)$.

Note that the line we want is parallel to $BD$, so the equation of the line can be written as $3x-4y+k=0$.

Now, since this line passes through the midpoint of the line segment $AO$, i.e. $(12/5,9/5)$, we have $$3\cdot\frac{12}{5}-4\cdot\frac 95+k=0\Rightarrow k=0.$$ So, the line we want is $3x-4y+0=0$.

Answer 2

Notice, the line through the mid-points of the sides $AB$ & $AD$ should be parallel to the diagonal $BD$ (having slope $3/4$) hence the equation of the desired line: $$3x-4y+c=0\tag 1$$ Now, in right $\triangle AOB$, the above line is parallel to the base $OB$ & passes through the mid-point of side $AB$ hence by using A-A-A similarity of triangles, the line should pass through the mid-point of line $AO$ joining the points $A(3, 1)$ & $O\left(\frac{9}{5}, \frac{13}{5}\right)$ whose mid-point is given as $$\left(\frac{3+\frac{9}{5}}{2}, \frac{1+\frac{13}{5}}{2}\right)\equiv \left(\frac{12}{5}, \frac{9}{5}\right)$$ now, substituting the coordinates of the point $\left(\frac{12}{5}, \frac{9}{5}\right)$ in (1), we get $$3\left(\frac{12}{5}\right)-4\left(\frac{9}{5}\right)+c=0$$ $$c=0$$ now, substituting value of $c=0$ in (1), we get equation of the required line:$$3x-4y=0$$ now, comparing the above equation with $ax+by+c=0$, we get $a=3, b=-4, c=0$ hence, $$\color{red}{|a+b+c|}=|3-4+0|=\color{blue}{1}$$