સમબાજુ ચતુષ્કોણ ABCD ની બાજુઓ રેખાઑ x - y + 2 | vedclass

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Question

Let the coordinate $A$ be $(0,c)$

Equations of the given lines are

$x-y+2=0$

$7x-y+3=0$

we know that the diagonals of the rhombus will be

Vedclass · Accepted Answer

$\frac{5}{2}$

Vedclass · Answer

Let the coordinate $A$ be $(0,c)$

Equations of the given lines are

$x-y+2=0$

$7x-y+3=0$

we know that the diagonals of the rhombus will be parallel to the angle bisectors of the two given lines; $y=x+2$ and $y=7x+3$

$	herefore $ equation of angle bisectors is given as:

$\frac{{x - y + 2}}{{\sqrt 2 }} =  \pm \frac{{7x - y + 3}}{{5\sqrt 2 }}$

$5x - 5y + 10 =  \pm \left( {7x - y + 3} ight)$

$	herefore $Parallel equation of the diagonals are $2x+4y-7=0$ and $12x-6y+13=0$

$	herefore $ slopes of diagonals are $\frac{{ - 1}}{2}$ and $2$.

Now, slope of the diagonal from $A(0,c)$ and passing through $P(1,2)$ is $(2-c)$

$	herefore 2 - c = 2 \Rightarrow c = 0$  (not possible)

$	herefore 2 - c = \frac{{ - 1}}{2} \Rightarrow c = \frac{5}{2}$

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