Question

7 soit les points a(4,6), b(-2,8), c(9,-1) et d(-3,-3). quelle est la distance entre les points milieux de ab et de cd? arrondis ta réponse au dixième près.

Explanation:

Step1: Find midpoint of AB

The midpoint formula for two points \((x_1,y_1,z_1)\) and \((x_2,y_2,z_2)\) is \(M = (\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2},\frac{z_1 + z_2}{2})\). For \(A(4,6)\) and \(B(-2,8)\) (assuming 2D, maybe z=0? Wait, the other points have z, maybe 3D. Wait, A(4,6) maybe A(4,6,0)? Wait, the problem says points A(4,6), B(-2,8), C(9, -1), D(-3, -3). Wait, maybe 2D? Wait, no, C and D have two coordinates? Wait, maybe typo, maybe A(4,6,0), B(-2,8,0), C(9,-1,0), D(-3,-3,0)? Or maybe 2D with x and y. Wait, the midpoint of AB: \(x = \frac{4 + (-2)}{2}=1\), \(y=\frac{6 + 8}{2}=7\), so midpoint \(M_{AB}=(1,7)\).

Step2: Find midpoint of CD

For \(C(9,-1)\) and \(D(-3,-3)\), midpoint \(M_{CD}=(\frac{9+(-3)}{2},\frac{-1+(-3)}{2})=(3,-2)\).

Step3: Calculate distance between \(M_{AB}(1,7)\) and \(M_{CD}(3,-2)\)

Distance formula: \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\). So \(x_1 = 1,y_1 = 7,x_2 = 3,y_2 = -2\). Then \((3 - 1)^2+( -2 - 7)^2=2^2+(-9)^2 = 4 + 81 = 85\). Then \(d=\sqrt{85}\approx9.2\) (wait, \(\sqrt{85}\approx9.2195\), rounded to tenth is 9.2? Wait, no, wait maybe 3D? Wait, the original points: A(4,6) – maybe A(4,6, z1), B(-2,8,z2), C(9,-1,z3), D(-3,-3,z4). Wait, maybe the problem has a typo, and A is (4,6,0), B(-2,8,0), C(9,-1,0), D(-3,-3,0). Then midpoints as above. Wait, but let's check again. Wait, the user's problem: "Quelle est la distance entre les points milieux de AB et de CD?" So midpoint of AB: \(M_1 = (\frac{4 + (-2)}{2},\frac{6 + 8}{2})=(1,7)\). Midpoint of CD: \(M_2 = (\frac{9 + (-3)}{2},\frac{-1 + (-3)}{2})=(3,-2)\). Then distance between \(M_1(1,7)\) and \(M_2(3,-2)\) is \(\sqrt{(3 - 1)^2+(-2 - 7)^2}=\sqrt{4 + 81}=\sqrt{85}\approx9.2\) (wait, \(\sqrt{85}\) is approximately 9.2195, so to the tenth, 9.2? Wait, no, 9.2195 rounded to tenth is 9.2? Wait, 9.2195: the tenths place is 2, hundredths is 1, so we keep 2. Wait, no, 9.2195: 9.2 when rounded to tenth? Wait, no, 9.2195 is 9.2 (tenth) because the hundredth digit is 1, which is less than 5. Wait, no, 9.2195: 9.2 (tenth place: 2, next digit 1 <5). Wait, but let's recalculate:

\((3 - 1)=2\), squared is 4. \((-2 - 7)=-9\), squared is 81. Sum is 85. \(\sqrt{85}\approx9.2195\), so to the tenth, 9.2.

Wait, but maybe the points are 3D? Let's check the original points: A(4,6) – maybe A(4,6, a), B(-2,8,b), C(9,-1,c), D(-3,-3,d). But the problem statement in the image: "Soit les points A(4,6), B(-2,8), C(9, -1) et D(-3, -3). Quelle est la distance entre les points milieux de AB et de CD? Arrondis ta réponse au dixième près." So 2D, x and y. So midpoints:

\(M_{AB}=(\frac{4 + (-2)}{2},\frac{6 + 8}{2})=(1,7)\)

\(M_{CD}=(\frac{9 + (-3)}{2},\frac{-1 + (-3)}{2})=(3,-2)\)

Distance: \(\sqrt{(3 - 1)^2 + (-2 - 7)^2}=\sqrt{4 + 81}=\sqrt{85}\approx9.2\) (wait, 9.2195, so 9.2 when rounded to tenth? Wait, 9.2195: the first decimal is 2, second is 1, so we don't round up. So 9.2.

Wait, but let's check the calculation again:

\(3 - 1 = 2\), \(2^2 = 4\)

\(-2 - 7 = -9\), \((-9)^2 = 81\)

Sum: 4 + 81 = 85

\(\sqrt{85} \approx 9.2195\), so to the nearest tenth, 9.2.

Answer:

\(9.2\)