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發問:
1.Let θbe the acute angle between the two lines L1 : y = 2x and L2 : y =3x. a)Find tanθ b)Find the equation of the line other than L2 which makes an angleθ with L1 and passes through the origin.2.The foot of the perpendicular from the vertex A(2,-3) of ΔABC to BC is... 顯示更多 1.Let θbe the acute angle between the two lines L1 : y = 2x and L2 : y =3x. a)Find tanθ b)Find the equation of the line other than L2 which makes an angleθ with L1 and passes through the origin. 2.The foot of the perpendicular from the vertex A(2,-3) of ΔABC to BC is (3,-8). a)Find the equation of BC. b)If B is (-2,-a) and ∠BAC = 90°, find the coordinates of C. 更新: 我知道個ANS,我係要步驟.
最佳解答:
Slope of line L1 = 2 = tan A Slope of line L2 = 3 = tan B A + m = B (where m stands for theta.) tan m = tan (B -A) = (tan B - tan A)/(1 + tan A tan B) = (3 -2)/(1 + 6) = 1/7. Let slope of the other line be tan C. so C + m = A tan C = tan (A - m) = (tan A - tan m)/(1 + tan A tan m) = (2 - 1/7)/(1 + 2/7) = (13/7)/(9/7) = 13/9. so the other line is y = 13x/9 or 9y = 13x. Q2. Let slope of BC be m, so m[(-8 +3)/(3-2)] = -1 -5m = -1, m = 1/5. so equation of BC is y + 8 = (x -3)/5 5y + 40 = x - 3 5y = x - 43. For point B, x - coordinate = -2 and it is on line BC, so 5(-a) = -2 - 43 = -45 a = 9, so coordinates of B is (-2,-9). Let coordinates of point C be (h,k) and it is on line BC, so 5k = h - 43................(1) Slope of AC x slope of AB = -1 [(k+3)/(h + 2)][(-9 + 3)/(-2 -2)] = -1 (k+3)/(h + 2) [-5/-4] = -1 5(k + 3) = -4(h +2) 5k + 15 = -4h - 8 5k + 4h = - 23.............(2) Sub. (1) into (2) we get 5k + 4(5k + 43) = - 23 25k = -23 - 172 = -195 k = -39/5 h = 5k + 43 = -39 + 43 = 4 so C is (4, -39/5) 2009-04-17 15:57:06 補充： Correction: slope of AC should be (k+3)/(h -2), not (k+3)/(h + 2). Slope of BC should be (-6/-4), not (-5/-4). So eqt (2) should be 3k + 2h = -5. Solving the 2 eqts, C should be (8,-7). Sorry for the mistake.
其他解答:
1. a) 1/7 b) y = 13/9 x 2. a) x - 5y - 43 = 0 b) (8, -7)
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A.mat,straight lines 快幫手.發問:
1.Let θbe the acute angle between the two lines L1 : y = 2x and L2 : y =3x. a)Find tanθ b)Find the equation of the line other than L2 which makes an angleθ with L1 and passes through the origin.2.The foot of the perpendicular from the vertex A(2,-3) of ΔABC to BC is... 顯示更多 1.Let θbe the acute angle between the two lines L1 : y = 2x and L2 : y =3x. a)Find tanθ b)Find the equation of the line other than L2 which makes an angleθ with L1 and passes through the origin. 2.The foot of the perpendicular from the vertex A(2,-3) of ΔABC to BC is (3,-8). a)Find the equation of BC. b)If B is (-2,-a) and ∠BAC = 90°, find the coordinates of C. 更新: 我知道個ANS,我係要步驟.
最佳解答:
Slope of line L1 = 2 = tan A Slope of line L2 = 3 = tan B A + m = B (where m stands for theta.) tan m = tan (B -A) = (tan B - tan A)/(1 + tan A tan B) = (3 -2)/(1 + 6) = 1/7. Let slope of the other line be tan C. so C + m = A tan C = tan (A - m) = (tan A - tan m)/(1 + tan A tan m) = (2 - 1/7)/(1 + 2/7) = (13/7)/(9/7) = 13/9. so the other line is y = 13x/9 or 9y = 13x. Q2. Let slope of BC be m, so m[(-8 +3)/(3-2)] = -1 -5m = -1, m = 1/5. so equation of BC is y + 8 = (x -3)/5 5y + 40 = x - 3 5y = x - 43. For point B, x - coordinate = -2 and it is on line BC, so 5(-a) = -2 - 43 = -45 a = 9, so coordinates of B is (-2,-9). Let coordinates of point C be (h,k) and it is on line BC, so 5k = h - 43................(1) Slope of AC x slope of AB = -1 [(k+3)/(h + 2)][(-9 + 3)/(-2 -2)] = -1 (k+3)/(h + 2) [-5/-4] = -1 5(k + 3) = -4(h +2) 5k + 15 = -4h - 8 5k + 4h = - 23.............(2) Sub. (1) into (2) we get 5k + 4(5k + 43) = - 23 25k = -23 - 172 = -195 k = -39/5 h = 5k + 43 = -39 + 43 = 4 so C is (4, -39/5) 2009-04-17 15:57:06 補充： Correction: slope of AC should be (k+3)/(h -2), not (k+3)/(h + 2). Slope of BC should be (-6/-4), not (-5/-4). So eqt (2) should be 3k + 2h = -5. Solving the 2 eqts, C should be (8,-7). Sorry for the mistake.
其他解答:
1. a) 1/7 b) y = 13/9 x 2. a) x - 5y - 43 = 0 b) (8, -7)
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