Find the eqution of line ,which is perpendicular 5x-2y=7 and passing through the mid point of the line joining (2,7)and(-4,1)
Answers
Answer:
first let us find out the coordinates of the mid point of the line joining (2,7) and (-4,1) using mid point formula and let this point be P.
P(x,y) = [(x1+x2)/2, (y1+y2)/2 ]
=> P (x,y) = [ (2-4)/2, (7+1)/2 ]
=> P(x,y) = (-1, 4)
as we have coordinates of P, to form an equation, we need to get the slope of this line.
since the line passing through P is perpendicular to the line 5x-2y=7, we can find the required slope by using the formula M1*M2 = -1, where M1 is the slope of the given line and M2 is the slope of the line we are supposed to form an equation for.
to find M1 , let us rewrite the given equation in y = M1X + C form.
5x - 2y = 7
=> - 2y = -5x + 7
=> y = -5x/-2 + 7/(-2)
=> y = 5/2 x - 7/2
on comparing this equation with y = M1X + c
we get M1 = 5/2
now using the equation M1*M2 = -1, we get
5/2 × M2 = -1
therefore M2 = -2/5
now as we know M2 and coordinates of P (-1,4) can use slope point form to get the equation
=> (y - y1) = M2(x - x1)
=> y - 4 = -2/5(x - (-1) )
=> y - 4 = -2/5( x + 1)
=> 5(y - 4) = -2 (x + 1) [ by cross multiplication ]
=> 5y - 20 = -2x -2
=> 2x + 5y - 18 = 0 is the answer