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  • The locus of the mid-points of the perpendiculars drawn from points on the line,x=2y to the line x=y is : Option: 1 Option: 2</strong

The locus of the mid-points of the perpendiculars drawn from points on the line,x=2y to the line x=y is :
Option: 1 2x-3y=0
Option: 2 3x-2y=0
Option: 3 5x-7y=0
Option: 4 7x-5y=0
 

Answers (1)

best_answer

 

 

Straight Line -

Straight Line

The slope of the line joining two Points

If\mathrm{A\left ( x_1,y_1 \right )} and \mathrm{B\left ( x_2,y_2 \right )} are two points on a straight line then the slope of the line is 

.\tan\theta=\frac{BC}{AC}=\frac{y_2-y_1}{x_2-x_1}

-

 

 

Line parallel and perpendicular to a given line -

Line parallel and perpendicular to a given line

The equation of the line parallel to ax + by + c = 0 is given as ax + by + λ = 0, where λ is some constant. 

Equation of the given line is ax + by + c = 0 

Its slope is (-a/b)

So, any equation of line parallel to ax + by + c = 0 is 

The equation of the line perpendicular to ax + by + c = 0 is given as bx - ay + λ = 0, where λ is some constant. 

Equation of the given line is ax + by + c = 0 

Its slope is (-a/b)

Slope of perpendicular line will be (b/a)    

So, any equation of line perpendicular to ax + by + c = 0 is 

-

 

 

\\\text { slope of } P Q =\frac{x-a}{y-2a}=-1 \\\\ \Rightarrow x-a=-y+2a \\\\ \Rightarrow a=\frac{x+y}{3}

Using midpoint
\\2x=2a+b\\2y=a+b

a=2x-2y

\frac{\mathrm{x}+\mathrm{y}}{3}=2(\mathrm{x}-\mathrm{y})

so locus is 6 x-6 y=x+y \quad \Rightarrow \quad 5 x=7 y

Correct Option (3)

Posted by

Ritika Jonwal

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