The diagonals of a rectangle pqrs intersect at 5,3 given that the equation for a line pq is 4y-9x=13 and that of line ps y-4x=5 what are coordinates of p and r ,what is the equation of line rq ,what is the equation of perpendicular line drawn to meet pr at 5,3 showing solution
The lines given for PQ and PS intersect at (-1,1) so that must mean that point P is at (-1,1)
If X = (5,3) then X-P = (6,2)
so R = X+(6,2) = (11,5)
PR has slope 1/3, so the perpendicular to PR has slope -3, making its equation
y-3 = -3(x-5)
Now, Q and S must lie on a circle of radius √40 with center at (5,3)
So the equation of the circle is (x-5)^2 + (y-3)^2 = 40
That circle intersects PQ and PS at Q and S, respectively.
The problem here is that the alleged lines PQ and PS do not intersect the circle anywhere near the points needed to make a rectangle.
To find the coordinates of points P and R, we need to find the intersection points of the given lines PQ and PS.
1. Finding point P:
Substitute the equation of line PQ (4y - 9x = 13) into the equation of line PS (y - 4x = 5) to solve for x.
y - 4x = 5
4y - 9x = 13
Multiply the second equation by 4 before subtracting:
4y - 16x = 20
4y - 9x = 13
------------
-7x = 7
Divide both sides by -7:
x = -1
Substitute this value of x back into the equation y - 4x = 5 to solve for y:
y - 4(-1) = 5
y + 4 = 5
y = 1
So, point P is at (-1, 1).
2. Finding point R:
To find point R, we need to find the intersection point of the diagonals, which is given as (5, 3).
Therefore, point R is at (5, 3).
3. Equation of line RQ:
The equation of the line RQ can be found using the slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept.
The slope of line RQ can be determined using the coordinates of points R and Q:
slope = (y2 - y1) / (x2 - x1) = (3 - 1) / (5 - (-1)) = 2 / 6 = 1/3
y = (1/3)x + b
To find b, substitute the coordinates of point R (x = 5, y = 3) into the equation:
3 = (1/3)(5) + b
3 = 5/3 + b
Multiply through by 3 to remove the fraction:
9 = 5 + 3b
Subtract 5 from both sides:
3b = 4
Divide both sides by 3:
b = 4/3
Therefore, the equation of line RQ is y = (1/3)x + 4/3.
4. Equation of perpendicular line through (5, 3):
To find the equation of a line perpendicular to RQ and passing through the point (5, 3), we need to determine the negative reciprocal of the slope of RQ.
The slope of RQ is 1/3, so the negative reciprocal is -3.
Using the point-slope form (y - y1) = m(x - x1), where (x1, y1) = (5, 3):
y - 3 = -3(x - 5)
y - 3 = -3x + 15
y = -3x + 18
Hence, the equation of the perpendicular line through (5, 3) is y = -3x + 18.
To find the coordinates of points P and R, we need to solve the system of equations formed by the equations of lines PQ and PS.
1. Equation of line PQ: 4y - 9x = 13
2. Equation of line PS: y - 4x = 5
To find the equation of line RQ, we need to use the fact that diagonals of a rectangle are perpendicular bisectors of each other. This means that the slope of line RQ is the negative reciprocal of the slope of line PQ. Let's start by finding the slope of line PQ.
1. Rearrange equation 1 to the slope-intercept form (y = mx + b):
4y - 9x = 13
4y = 9x + 13
y = 9/4x + 13/4
The slope of line PQ is 9/4.
To find the equation of the perpendicular line, we need to find its slope. The slope of the perpendicular line will be the negative reciprocal of the slope of line PQ. Let's calculate it:
1. Calculate the negative reciprocal of the slope of line PQ:
Negative reciprocal = -1 / (9/4)
Negative reciprocal = -4/9
Now we have the slope of the perpendicular line, -4/9.
Since we know a point on the line (5,3), we can use the point-slope form of a line to find the equation of the perpendicular line. The equation is given by:
y - y₁ = m(x - x₁)
Where (x₁, y₁) is the given point (5,3) and m is the slope of the perpendicular line (-4/9).
2. Substitute the values into the point-slope form equation:
y - 3 = (-4/9)(x - 5)
Simplifying this equation will give us the final equation of the perpendicular line.
To find the coordinates of points P and R, we can find the intersection point of lines PQ and PS.
3. Substitute the equation of line PS (y - 4x = 5) into equation 1 and solve for x:
4(y - 4x) = 5
4y - 16x = 5
Substitute this value of 4y - 16x into equation 1:
4y - 16x = 9x + 13
Simplify and solve for x:
4y - 25x = 13
Rearrange to solve for y:
4y = 25x + 13
y = 25/4x + 13/4
Now we have two expressions for y in terms of x:
y = 9/4x + 13/4 (from equation 1)
y = 25/4x + 13/4 (from equation 3)
Set them equal to each other and solve for x:
9/4x + 13/4 = 25/4x + 13/4
Simplify and solve for x:
9x - 25x = 13 - 13
-16x = 0
x = 0
Substitute this value of x into either equation to solve for y:
y = 9/4(0) + 13/4
y = 13/4
Therefore, coordinates of point P are (0, 13/4).
Repeat the process for R by substituting the x-coordinate (0) into either equation to solve for y:
y = 25/4(0) + 13/4
y = 13/4
Therefore, coordinates of point R are (0, 13/4).
Therefore, point P is located at (0, 13/4) and point R is also located at (0, 13/4).
The equation of line RQ can be found by using the point-slope form of a line. We already have a point on the line, which is (0, 13/4), and we know the slope of the line is the negative reciprocal of the slope of line PQ, which is -4/9. Using the point-slope form, the equation of line RQ is:
y - 13/4 = -4/9(x - 0)
y - 13/4 = -4/9x
Simplifying this equation will give us the final equation of line RQ.