  Question: A Maths question based on straight lines Posted by: Tejas k. on 28.10.2020  • Like

The equations of the cross roads where the elephant is standing are

x- y+2=0    ………  A

y- 1=o.     …………B   that is y= 1

putting y= 1   in equation A we get x-1+2=0 or x=  (-)1

Therefore the co ordinates of the junction where elephant is standing is   (-)1, 1.

Given the equation of the road where the elephant wants to go is x - y-3=0 or y = x-3.

The shortest distance of a point to a given line is the perpendicular distance of the point to the given line. The slope of a line perpendicular to a given line  will be equal to the negative reciprocal of the given line. That is if the slope of a line is “m “, then the slope of the line perpendicular to this line will be “(-1/m)”.

Equation  of the line representing the road where the elephant wants to go is  y=x-3.Its slope = 1.

Therefore the slope of the  line perpendicular to the above line is (-)1.Therfore the general equation of the perpendicular is   y=  - x +k, where ‘k” is a constant        …….C .

Now the junction (where the elephant is standing) is a point on the perpendicular whose co ordinates we have found as (-)1 , 1.Therfore this point should satisfy the equation of the perpendicular. Putting these values in equation “C” ,we get  1= -{ (-)1} +k = 1+k which gives the value of k=0.

Therefore the equation of the perpendicular is y= - x+0   or  x +y=0.

Equation of the path to be followed by the elephant is x+ y=0

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