# maths

P(6,3), Q(3,7), and R(4,2) are three points in a plane. A is the midpoint of QR and B is the foot of the perpendicular from Q to PR. Find;-
(A) the cordinates of A;
(B) the equations of the lines PA and QB;
(C) the point of intersection of the lines PA and QB;
(D) the equation of the line passing through Q and parallel to the line PR.

need help I don't understand it

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1. Midpoint is (x1 + x2)/2, (y1 + y2)/2
Label your coordinates (x1, y1) and (x2, y2)
and begin : )

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2. Plot the given points on a Cartesian plane, label them, and join them with a ruler. This helps immensely to see
what needs to be done.
a) Find the midpoint A of QR using the formula Ms Pi provided. Use points P and Q.
b) PA is a median and QR is an altitude. Rough sketch them in. To get the equations of them, use what you know about slopes as well as y = mx + b.
c) Use substitution or elimination to solve the system of two equations you found in part b).
d) Rough sketch this line and use what you know about slopes as well as y = mx + b to get its equation.

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3. Correction on item a):
Use points Q and R :)

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4. Using the method that Ms Pi told you,
A is ( (4+3)/2 , (2+7)/2 = A( 7/2,9/2)

for equation of PA
slope PA = (9/2-3)/(7/2-6) = -3/5
so equation: y-3 = (-3/5)(x-6)
5y - 15 = -3x + 18
3x + 5y = 33 <--- equation of AP

slope of PR = (3-2)/(6-4) = 1/2
slope slope of QB = -2
equation of QB: y-7 = -2(x-3)
y-7 = -2x + 6
2x + y = 13 or y = 13-2x

sub into 3x + 5y = 33
3x + 5(13-2x) = 33
3x + 65 - 10x = 33
-7x = - 32
x = 32/7
back into y = 13 - 2x
y = 27/7

The last is straightforward, you do it, let me know what you get

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Reiny

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