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please help!
hey guys i'm quite stuck and could do with some help:
A triangle has vertices P(-2,2), Q (q,0) and R (5,3).
a) The side PQ is twice as long as side QR. find the possible values of q.
b) show that triangle PQR is right-angled when q = 4
c) when q = 4 find the coordinates of the centre of the circle which passes through P, Q, R.
thanks a lot!
A triangle has vertices P(-2,2), Q (q,0) and R (5,3).
a) The side PQ is twice as long as side QR. find the possible values of q.
b) show that triangle PQR is right-angled when q = 4
c) when q = 4 find the coordinates of the centre of the circle which passes through P, Q, R.
thanks a lot!
a) PQ = √[(q--2)² + (0-2)²]
QR = √[(5-q)²+(3-0)²]
PQ = 2QR
√[(q--2)² + (0-2)²] = 2 x √[(5-q)²+(3-0)²]
{√[(q--2)² + (0-2)²]}² = {2 x √[(5-q)²+(3-0)²]}²
(q--2)² + (0-2)² = 4 x [(5-q)²+(3-0)²]
q² + 4q + 4 + 4 = 4 [ 25 - 10q + q² + 9]
solve this and hence u got the possible values of q.
b) tangent of PQ = (0-2)/(4--2) = -1/3
tangent of QR = (3-0)/(5-4) = 3
-1/3 x 3 = -1 ==> PQ and QR are perpendicular ==> right-angled triangle!
QR = √[(5-q)²+(3-0)²]
PQ = 2QR
√[(q--2)² + (0-2)²] = 2 x √[(5-q)²+(3-0)²]
{√[(q--2)² + (0-2)²]}² = {2 x √[(5-q)²+(3-0)²]}²
(q--2)² + (0-2)² = 4 x [(5-q)²+(3-0)²]
q² + 4q + 4 + 4 = 4 [ 25 - 10q + q² + 9]
solve this and hence u got the possible values of q.
b) tangent of PQ = (0-2)/(4--2) = -1/3
tangent of QR = (3-0)/(5-4) = 3
-1/3 x 3 = -1 ==> PQ and QR are perpendicular ==> right-angled triangle!
cinders1288
hey guys i'm quite stuck and could do with some help:
A triangle has vertices P(-2,2), Q (q,0) and R (5,3).
a) The side PQ is twice as long as side QR. find the possible values of q.
b) show that triangle PQR is right-angled when q = 4
c) when q = 4 find the coordinates of the centre of the circle which passes through P, Q, R.
thanks a lot!
A triangle has vertices P(-2,2), Q (q,0) and R (5,3).
a) The side PQ is twice as long as side QR. find the possible values of q.
b) show that triangle PQR is right-angled when q = 4
c) when q = 4 find the coordinates of the centre of the circle which passes through P, Q, R.
thanks a lot!
(a)
|PQ| = √[(q+2)² + (-2)²]
|QR| = √[(q-5)² + (-3)²]
mod(PQ) = 2mod(QR)
So: √[(q+2)² + (-2)²] = 2√[(q-5)² + (-3)²]
Therefore: (q+2)² + (-2)² = 4[(q-5)² + (-3)²]
q² + 4q + 4 + 4 = 4[q² - 10q + 25 + 9]
q² + 4q + 8 = 4q² - 40q + 136
3q² - 44q + 128 = 0
Check the discriminant: (-44)² - 4(3)(128) = 1936 - 1536 = 400
Therefore it factorises.
3q² - 44q + 128 = 0
(3q - 32) (q - 4 ) = 0
q = 32/3 or q = 4
(b)
When q = 4: P(-2,2), Q (4,0) and R (5,3)
m = (y2 - y1)/(x2 - x1)
mPQ = -2/6 = -1/3
mQR = 3/1 = 3
Perpendicular gradients have product -1.
mPQ x mQR = -1/3 x 3 = -1
So triangle PQR has a right angle.
(c)
Circle passes through P, Q and R. The method to use to find the coordinates of the center will use the following idea. Two perpendicular bisectors (of different chords) intersect at the centre of a circle. This method is fairly long.
Firstly: Mid PointPQ(1,1)
mPQ = -1/3.
Gradient of perpendicular bisector is -1/(-1/3) = 3
Equation of perpendicular bisector is (y - y1) = m(x - x1)
So: y - 1 = 3x - 3
so: y = 3x - 2
Oo! I've noticed a shortcut. There is a cirecle theorem which states the angle at the centre is twice the angle at the circumference. In a special case this gives the "right angle in a semicircle" rule. We have a right angle at angle PQR as PQ and QR are perpendicular, so the line PR is the diameter of the circle. Therefore PR passes through the centre of the circle. This can be used as the second equation.
mPR = 1/7.
Equation is: (y - y1) = m(x - x1)
So: y - 3 = 1/7(x - 5)
so: y = x/7 + 16/7
Equate: 3x - 2 = x/7 + 16/7
so: 21x - 14 = x + 16
20x = 30
x = 1.5
y = 3(1.5) - 2 = 2.5
Therefore centre of circle is: (1.5, 2.5)
---------
To be honest I'd be fairly surprised if I hadn't made a few mistakes in all that working. Sorry!
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