# the lones x-2y+6=0 and 2x-y-10+0 intersect at P.Without finding the coordinates of prove that the equation of the line through P and the origin of coordinates is perpendicular to 39x+33y-580=0.

First write the equation of line in the form of   ,  now slope of the line is

Slope of line perpendicular to the above line is-

Equation of line passing through the origin and with slope equal to 11/13 is given by-

Substituting in , we get

Substituting in , we get

Since the point of intersection of lines ,  and is same. The line passes through P and origin is perpendicular to .

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assuming that the line is perpendicular to 39x+33y-580=0

so the slope of that line will be

m*(-39/33) = -1

m= 33/39 =11/13

As this line is passing thru' the origin the equation is y =mx or y= 11/13x

Now this line is intersecting with two lines x-2y+6=0 and 2x-y-10=0

Therefore first we substitute the value of y in equation 1 , we get some value for x

Now we substitute the value of y in equation 2 , we get some value for x

But as all these three lines intersect at the same point the values of x must be equal  , and if they are equal then we proved that the line is perpendicular to 39x+33y-580=0 as it is satisfying all the conditions..

I hope u got it..

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