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
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
Midpoint is (x1 + x2)/2, (y1 + y2)/2
Label your coordinates (x1, y1) and (x2, y2)
and begin : )
The last question for me is a lil bit tricky so can I get the answer 😊😊😊😁😅
Correction on item a):
Use points Q and R :)
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.
to right pqrs is arectangle if ps = 5cm and pr =13cm find sr and qs
How did you get the slope of QB and the lesson was great
Pls help me find the equation of the line passing through Q and parallel to the line PR
Sure, I'd be happy to help you understand how to solve this problem step by step.
(A) To find the coordinates of point A, the midpoint of QR, we can use the midpoint formula. The midpoint formula states that the coordinates of the midpoint (A) of a line segment with endpoints (x1, y1) and (x2, y2) are given by the average of the coordinates:
xA = (x1 + x2)/2
yA = (y1 + y2)/2
In this case, the coordinates of Q are (3, 7) and the coordinates of R are (4, 2). Plugging these values into the midpoint formula, we get:
xA = (3 + 4)/2 = 7/2
yA = (7 + 2)/2 = 9/2
So the coordinates of point A are (7/2, 9/2).
(B) To find the equation of line PA, we need to determine the slope of PA and use the point-slope form of the equation of a line. The slope of a line passing through two points (x1, y1) and (x2, y2) is given by:
m = (y2 - y1)/(x2 - x1)
In this case, we know the coordinates of A (7/2, 9/2) and P (6, 3). Plugging these values into the slope formula, we get:
m = (9/2 - 3)/(7/2 - 6) = (3/2)/(7/2 - 12/2) = (3/2)/(-5/2) = -3/5
Now, we can use the point-slope form of the equation of a line, which states that a line passing through a point (x1, y1) with slope m is given by:
y - y1 = m(x - x1)
Plugging in the values from point P (6, 3) and the slope -3/5, we get:
y - 3 = (-3/5)(x - 6)
Simplifying this equation, we get the equation of line PA.
Similarly, to find the equation of line QB, we need to determine its slope and use the point-slope form. The slope of QB can be found using the coordinates of Q (3, 7) and B. However, we do not yet know the coordinates of B. So, we need to find B first.
To find the coordinates of B, the foot of the perpendicular from Q to PR, we can use the fact that the slope of a line perpendicular to another line is the negative reciprocal of the slope of the original line. In other words, if the slope of PR is m, then the slope of the line perpendicular to PR is -1/m.
The slope of PR can be found using the coordinates of P (6, 3) and R (4, 2) using the slope formula mentioned earlier. Once we have the slope of PR, we can find the slope of QB by taking the negative reciprocal of that value.
(C) The point of intersection of the lines PA and QB can be found by solving the system of equations formed by the equations of PA and QB. This involves solving for the values of x and y that satisfy both equations simultaneously.
(D) To find the equation of the line passing through Q and parallel to PR, we can use the fact that parallel lines have the same slope. We have already found the slope of PR in part (B), so we can use that slope to find the equation of the line passing through Q.