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
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-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.