(a) In the diagram, MNPQ is a circle with centre O, |MN| = |NP| and < OMN = 50°. Find: (I) < MNP (ii) < POQ (b) Find the equation of the line which has the ...

(a) In the diagram, MNPQ is a circle with centre O, |MN| = |NP| and < OMN =  50°. Find:

(I) < MNP

(ii) < POQ

 (b) Find the equation of the line which has the same gradient as 8y + 4xy = 24 and passes through the point (-8, 12)

Answer Details

(a)

1. Since |MN| = |NP|, then ∠MNP = ∠NMP (angles opposite equal sides of an isosceles triangle are equal). Therefore, ∠MNP = (180 - ∠OMN)/2 = (180 - 50)/2 = 65 degrees.
2. ∠POQ is the angle subtended at the center of the circle by arc MP. Since MN = NP, then arc MN = arc NP. Therefore, arc MP = arc MN + arc NP = 2arc MN. So, ∠POQ = (1/2)arc MP = (1/2)(2arc MN) = arc MN. Since ∠OMN = 50 degrees, then arc MN = 2∠OMN = 100 degrees. Therefore, ∠POQ = arc MN = 100 degrees.

(b) To find the equation of the line which has the same gradient as 8y + 4xy = 24 and passes through the point (-8, 12), we need to rewrite the given equation in slope-intercept form, y = mx + c, where m is the gradient and c is the y-intercept.

8y + 4xy = 24
Factor out y on the left side:
y(8 + 4x) = 24
Divide both sides by (8 + 4x):
y = 3/(2 + x)

So the gradient of the line is 3. Since the line passes through (-8, 12), we can use the point-slope form of the equation of a straight line to find the equation of the line:

y - y1 = m(x - x1)
y - 12 = 3(x + 8)
y - 12 = 3x + 24
y = 3x + 36

Therefore, the equation of the line which has the same gradient as 8y + 4xy = 24 and passes through the point (-8, 12) is y = 3x + 36.