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C2 Coordinate geometry
The points P (−10, 2), Q (8, 14) and R (−2, −10) all lie on circle C.
a) Show that PR is perpendicular to PQ.
b) Hence, show that C has the equation x2 + y2 − 6x − 4y − 156 = 0.
a) Show that PR is perpendicular to PQ.
b) Hence, show that C has the equation x2 + y2 − 6x − 4y − 156 = 0.
Reply 1
14
firstly findf the equation of Pr and then find the normal of it, as for te circle equate them all
Reply 2
You don't need to find the equation of PR. A formula for the gradient of a straight line is (y1-y2)/(x1-x2).
The gradient of PR: (2--10)/(-10--2) = 12/-8 = -3/2.
The gradient of a perpendicular line to this would be 2/3.
The gradient of PQ: (2-14)/(-10-8) = -12/-18=2/3.
So PR and PQ are perpendicular.
The gradient of PR: (2--10)/(-10--2) = 12/-8 = -3/2.
The gradient of a perpendicular line to this would be 2/3.
The gradient of PQ: (2-14)/(-10-8) = -12/-18=2/3.
So PR and PQ are perpendicular.
Reply 3
Fenchurch
The points P (−10, 2), Q (8, 14) and R (−2, −10) all lie on circle C.
b) Hence, show that C has the equation x2 + y2 − 6x − 4y − 156 = 0.
b) Hence, show that C has the equation x2 + y2 − 6x − 4y − 156 = 0.
1.) Find an equation of a line going through the mid point of perpendicular to PQ.
2.) Find an equation of a line going through the mid point of perpendicular to PR.
3.) Find where they intersect.
1.) y - (8) = -3/2(x - (-1))
y = -3/2x + 13/2
2.) y - (-4) = 2/3(x - (-6))
y = 2/3x + 8
3.) 2/3x + 8 = -3/2x + 13/2
4/6x + 42/6 = -9/6x + 39/6
4x + 42 = -9x + 39
13x = -3, x = -3/13
Thats my attempt, I wouldn't rely on that too much as I always seem to make some silly mistakes. I'm sure you could finish it off.
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