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In the coordinate plane, the points A = (-6,2) , B = (-3, 3) , C = (0,n) and D = (m,0) form a quadrilateral ABCD. Find the value of n so that the perimeter of the quadrilateral ABCD is the least. 更新: 在坐標平面上，點 A = (-6,2) 、B = (-3, 3) 、 C = (0,n) 及 D = (m,0) 組成一個四邊形ABCD。求 n 的值使得該四邊形 ABCD 的周界為最短。
最佳解答:
We first check what the possible values of n there are. Slope of line segment BC = (3-n)/(-3-0)=(3-n)/(-3) Slope of line segment AD = (2-0)/(-6-m)=(2)/(-6-m) Since ABCD is a quadrilateral, slope of line segment BC=slope of line segment AD, therefore, (3-n)/(-3)= (2)/(-6-m), or (3-n)(6+m)=6 ---------------- (1) Similarly, slope of line segment BA=slope of line segment CD, therefore, (3-2)/(-3+6)=(n)/(-m) 1/3=(n)/(-m) m=-3n ----------- (2) Subs. (2) into (1), we have (3-n)(6-3n)=6 3(3-n)(2-n)=6 (3-n)(2-n)=2 n^2-5n+6=2 n^2-5n+4=0 (n-1)(n-4)=0 n=1 or n=4 When n=4, m=-12, C=(0,4) and D=(-12,0), then points A, B, C and D are on one single straight line and hence ABCD cannot be a quadrilateral. Hence n=4 is rejected. When n=1, m=-3, C=(0,1) and D=(-3,0). ABCD is a quadrilateral. Since there is one possible value of n (n=1), this value of n is the value with which the perimeter of the quadrilateral ABCD is the least. Hope this is of assistance to you! Happy New Year!
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發問:In the coordinate plane, the points A = (-6,2) , B = (-3, 3) , C = (0,n) and D = (m,0) form a quadrilateral ABCD. Find the value of n so that the perimeter of the quadrilateral ABCD is the least. 更新: 在坐標平面上，點 A = (-6,2) 、B = (-3, 3) 、 C = (0,n) 及 D = (m,0) 組成一個四邊形ABCD。求 n 的值使得該四邊形 ABCD 的周界為最短。
最佳解答:
We first check what the possible values of n there are. Slope of line segment BC = (3-n)/(-3-0)=(3-n)/(-3) Slope of line segment AD = (2-0)/(-6-m)=(2)/(-6-m) Since ABCD is a quadrilateral, slope of line segment BC=slope of line segment AD, therefore, (3-n)/(-3)= (2)/(-6-m), or (3-n)(6+m)=6 ---------------- (1) Similarly, slope of line segment BA=slope of line segment CD, therefore, (3-2)/(-3+6)=(n)/(-m) 1/3=(n)/(-m) m=-3n ----------- (2) Subs. (2) into (1), we have (3-n)(6-3n)=6 3(3-n)(2-n)=6 (3-n)(2-n)=2 n^2-5n+6=2 n^2-5n+4=0 (n-1)(n-4)=0 n=1 or n=4 When n=4, m=-12, C=(0,4) and D=(-12,0), then points A, B, C and D are on one single straight line and hence ABCD cannot be a quadrilateral. Hence n=4 is rejected. When n=1, m=-3, C=(0,1) and D=(-3,0). ABCD is a quadrilateral. Since there is one possible value of n (n=1), this value of n is the value with which the perimeter of the quadrilateral ABCD is the least. Hope this is of assistance to you! Happy New Year!
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