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સમબાજુ ચતુષ્કોણ $ABCD$ ની બાજુઓ રેખાઑ $x - y + 2\, = 0$ અને $7x - y + 3\, = 0$ ને સમાંતર છે. જો સમબાજુ ચતુષ્કોણના વિકર્ણો બિંદુ $P( 1, 2)$ આગળ છેદે અને શિરોબિંદુ $A$ ( ઉંગમબિંદુથી અલગ) એ $y$ અક્ષ પર આવેલ છે $A$ નો $x-$ યામ મેળવો.
$2$
$\frac{7}{4}$
$\frac{7}{2}$
$\frac{5}{2}$
Solution
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 $y=7x+3$
$\therefore $ equation of angle bisectors is given as:
$\frac{{x – y + 2}}{{\sqrt 2 }} = \pm \frac{{7x – y + 3}}{{5\sqrt 2 }}$
$5x – 5y + 10 = \pm \left( {7x – y + 3} \right)$
$\therefore $Parallel equation of the diagonals are $2x+4y-7=0$ and $12x-6y+13=0$
$\therefore $ 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 $(2-c)$
$\therefore 2 – c = 2 \Rightarrow c = 0$ (not possible)
$\therefore 2 – c = \frac{{ – 1}}{2} \Rightarrow c = \frac{5}{2}$
