(a) The graph of (y = 2px^{2} – p^{2}x – 14) passes through the point (3, 10). Find the values of p.
(b) Two lines, (3y – 2x = 21) and (4y + 5x = 5) intersect at the point Q. Find the coordinates of Q.
Explanation
(a) (y = 2px^{2} – p^{2}x – 14)
At point (3, 10), y = 10 when x = 3.
(implies 10 = 2p(3^{2}) – p^{2}(3) – 14)
(10 = 18p – 3p^{2} – 14)
(-10 = 3p^{2} – 18p + 14 implies 3p^{2} – 18p + 24 = 0)
(3p^{2} – 12p – 6p + 24 = 0)
((p – 4)(3p – 6) = 0 implies text{p = 2 or 4}).
(b) (3y – 2x = 21 … (1))
(4y + 5x = 5 ….(2))
Using elimination method, multiply (1) by 4 and (2) by 3.
((1) times 4 : 12y – 8x = 84 … (3))
((2) times 3 : 12y + 15x = 15 … (4))
Subtracting (3) – (4), we have:
(-23x = 69 implies x = -3)
Putting x = -3 in (1), we have
(3y – 2(-3) = 3y + 6 = 21)
(3y = 15 implies y = 5)
Hence, the coordinates of Q are (-3, 5).