✔ 最佳答案
Q1. Let B be the origin (0,0), so A is (0,15), C is (19,0) and P is (10,8). Using the formula of distance between 2 points, you can find the length of AP, BP and CP. The one with the shortest distance will be the answer.
Q2.
(a) Using the formula for slope, slope of BC = (k - 7)/(k - 2) which is equal to 2/3. So (k-7)/(k-2) = 2/3, 3(k -7) = 2(k-2), 3k - 21 = 2k - 4, so
k = 17, so C is (17,17).
So slope of AC = (17 - 2)/(17 + 3) = 15/20 = 3/4.
Slope of AB = tan 45 = 1.
So AB is having the greatest slope.
Q3.
Slope of LM = (-1 -3)/(-6 -4) = -4/-10 = 2/5.
Slope of MN = (-1-1)/(-6 + 1) = -2/-5 = 2/5.
since slope of LM = slope of MN, so L,M, and N are on a straight line.
Q4.
Since C is on the x -axis, the y-coordinate of C will be 0. Using the point of division formula, we get
0 = [(-3)(r) + (9)(1)]/(1 + r)
0 = (-3r + 9)/(1 + r), so r = 3.
so the x- coordinate of C = [(1)(3) + (9)(1)]/(1 + 3) = 12/4 = 3.
C is ( 3,0).
Q5.
(a) Let OP : OQ = 1 : r.
so 0 = [(-6)(r) + (9)(1)]/(1 + r)
0 = (-6r + 9)/( 1 + r), so r = 9/6 = 3/2.
That is OP : OQ = 1 : 3/2 or = 2 : 3.
So OA : AQ = 2 : 3.
The x - coordinate of A = [(0)(3) + (9)(2)]/( 2 + 3) = 18/5.
The y-coordinate of A = [(6)(2) + (0)(3)]/( 2 + 3) = 12/5.
so A is (18/5, 12/5).
Q6.
Let B is (0,0), C is ( c ,0) and A is ( a, b).
Coordinates of P is [3a/5, 3b/5].
Coordinates of Q is [(2c + 3a)/5, 3b/5].
Slope of BC = (0-0)/(c - 0) = 0.
Slope of PQ = [3b/5 - 3b/5]/[3a/5 - (2c + 3a)/5] = 0.
so BC//PQ.
Length of BC = c.
Length of PQ = sqrt {[ 3a/5 - (2c + 3a)/5]^2 + [3b/5 - 3b/5]^2}
= sqrt[(2c/5)^2] = 2c/5 = 2BC/5.
Note: You can assume A, B and C to be any coordinates, the result is the same, but calculation will be much more complicated.
