જો ત્રિકોણની બાજુઓ y = mx + a, y = nx + b અને | vedclass

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Question

Given lies are

$L_1: y=m x+a$

$L_2: y=n x+b$

Line $L_1$ intersects $y-$ axis at $A=(0, a)$

Line $L_2$ intersects $y-$ axis at $B=(0, b)$

Vedclass · Accepted Answer

$\frac{{1\,\,{{(a - b)}^2}}}{{2\,\,(m - n)}}$

Vedclass · Answer

Given lies are

$L_1: y=m x+a$

$L_2: y=n x+b$

Line $L_1$ intersects $y-$ axis at $A=(0, a)$

Line $L_2$ intersects $y-$ axis at $B=(0, b)$

Comparing the given two equations,

$m x+a=n x+b$

$\Rightarrow n x-m x=a-b$

$\Rightarrow x=\frac{a-b}{n-m}$

Substituting above value in $L_1$ we get,

$y=m \frac{a-b}{n-m}+a$

$\Rightarrow y=\frac{m a-m b+n a-m a}{n-m}$

$\Rightarrow y=\frac{n a-m b}{n-m}$

$	herefore$ The co-ordinates of intersection of the lines is $C =\left(\frac{a-b}{ n - m }, \frac{ na - mb }{ n - m }ight)$

Distance between $A$ and $B$

$d=\sqrt{(0-b)^2+(a-b)^2}$

$\Rightarrow d=\sqrt{(a-b)^2}$

$\Rightarrow d=a-b$

Length of perpendicular from $C$ on the $y$ - axis.

$p=\frac{a-b}{n-m}$

$	herefore$ Area of the triangle fromed by the lines $L _1, L _2$ and $y -$ axis

$A=\left|\frac{1}{2} d pight|$

$\Rightarrow A=\frac{1}{2} 	imes(a-b) 	imes \frac{a-b}{n-m}$

જો ત્રિકોણની બાજુઓ $y = mx + a, y = nx + b$ અને $x = 0,$ હોય, તો તેનું ક્ષેત્રફળ :

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