# Identify the coordinates of Q and R using the midpoint formula and creating two equations for the following case:There are two equations: x - 2y - 4 = 0 and x + y = 5 There is one point: P(1,1) A...

Identify the coordinates of Q and R using the midpoint formula and creating two equations for the following case:

There are two equations: x - 2y - 4 = 0 and x + y = 5

There is one point: P(1,1)

A line is drawn from P to intersect with x - 2y - 4 = 0 at Q and x + y = 5 at R, so that P is the midpoint of QR.

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Let the line ax+ by + c = 0 join the points Q and R and pass through P.

Let ax + by + c intersect x - 2y-4 = 0 at Q(x1, y1). Let ax + by + c intersect x +y - 5 = 0 at R(x2, y2).

As P is equidistant from these points (x1 + x2)/2 = 1 and (y1 + y2)/2 = 1

So we have x1 - 2y1 - 4 = 0, x2 + y2 - 5 = 0 and x1 + x2 = 2 and y1 + y2 = 2.

We can write x1 - 2y1 - 4 = 0 and 2 - x1 + 2 - y1 - 5 = 0

=> x1 - 2y1 - 4 = 0 and x1 + y1 + 1 =0

subtracting the two

=> 3y1 + 5 = 0

=> y1 = -5/3

x1 = -1 + 5/3 = 2/3

x2 = 2 - 2/3 = 4/3

y2 = 2 + 5/3 = 11/3

**Therefore Q is (2/3 , -5/3) and R is (4/3 , 11/3)**

A point P is given with coordinates P(1,1).

P(1,1) is the mid point of Q(x1, y1) and R(x2,y2), where Q is on x-2y-4 and R is on x+y = 5.

Therefore by mid point formula (x1+x2)/2 = 1 and (y1+y2)/2 = 1

x1+x2 = 2 ..(1)and y1+y2 = 2 ..(2).

Since (x1, y1) lies on x-2y-4 = 0, or x-2y = 4 and (x2,y2) lies on x+y = 5, we have:

x1-2y1 = 4 ...(3) and x2+y2 = 5....(4).

We solve the system of equations (1), (2) (3) and (4):

(1)-(3): x2+2y1 = -2... (5).

(4)-(2): x2-y1 = 3.........(6)

(5)-(6): 3y1 = -5. So y1 = -5/3. Put y1 = -5/3 in (6), and we get x2 = 3+y1 = 3-5/3 = 4/3.

Put x2 = 4/3 in (1), x1+x2 = 2. So x1 =2-x2 = 2-4/3 = 2/3.

Put y1 = -5/3 in (2): y1+y2 = 2. So y1 = 2-y2 = 2-(-5/3) = 11/3.

Therefore Q(x1,y1) = Q(2/3, -5/3) which lies on x-2y-4 = 0 and R(x2,y2) = (4/3, 11/3) which lies on x+y = 5.