Question

jefferson is plotting the vertices of an isosceles right triangle on a coordinate graph. he has plotted the two points shown. where should he plot the third point? $(-2,-9)$ $(4,2)$ $(-2,0)$ $(3,1)$

Explanation:

Step1: Identify the two plotted points

From the graph, the two points are \((-2, -4)\) and \((4, -4)\).

Step2: Analyze the distance between the two points

The distance between \((-2, -4)\) and \((4, -4)\) is calculated using the distance formula for horizontal points (since \(y\)-coordinates are the same). The distance \(d = |x_2 - x_1| = |4 - (-2)| = 6\).

Step3: Determine the properties of an isosceles right triangle

In an isosceles right triangle, the legs are equal in length, and the right angle is between them. So, the third point should form a right angle with the two given points, and the distance from the third point to each of the given points should be equal (or one leg is the horizontal distance we found, and the other leg is vertical with the same length).

Step4: Check each option

- For \((-2, -9)\): Distance from \((-2, -4)\) is \(|-9 - (-4)| = 5\), not 6. Distance from \((4, -4)\) is \(\sqrt{(4 - (-2))^2 + (-9 - (-4))^2}=\sqrt{36 + 25}=\sqrt{61} eq6\).
- For \((4, 2)\): Distance from \((-2, -4)\) is \(\sqrt{(4 - (-2))^2 + (2 - (-4))^2}=\sqrt{36 + 36}=\sqrt{72} eq6\). Distance from \((4, -4)\) is \(|2 - (-4)| = 6\). But the triangle formed would not be isosceles right (legs would be 6 and \(\sqrt{72}\)).
- For \((-2, 0)\): Distance from \((-2, -4)\) is \(|0 - (-4)| = 4\), not 6. Distance from \((4, -4)\) is \(\sqrt{(4 - (-2))^2 + (0 - (-4))^2}=\sqrt{36 + 16}=\sqrt{52} eq6\). Wait, correction: Wait, the two given points are \((-2, -4)\) and \((4, -4)\). The horizontal distance is 6. For an isosceles right triangle, if we take the vertical leg from \((-2, -4)\) upwards, adding 6 to the \(y\)-coordinate: \(-4 + 6 = 2\)? No, wait, the \(y\)-coordinate difference should be equal to the horizontal difference. Wait, the two points are on the line \(y = -4\). The midpoint is \((1, -4)\). The third point should be such that either:
- It has the same \(x\)-coordinate as one of the points and the \(y\)-coordinate differs by 6 (since the horizontal distance is 6). So, for \((-2, -4)\), moving up 6: \(y = -4 + 6 = 2\)? No, \((-2, 2)\) is not an option. Wait, moving down 6: \(y = -4 - 6 = -10\), not an option. For \((4, -4)\), moving up 6: \(y = -4 + 6 = 2\), so \((4, 2)\) – but earlier calculation was wrong. Wait, distance from \((-2, -4)\) to \((4, 2)\): \(\sqrt{(4 - (-2))^2 + (2 - (-4))^2}=\sqrt{36 + 36}=\sqrt{72}=6\sqrt{2}\), and distance from \((4, -4)\) to \((4, 2)\) is 6. So that's a right triangle with legs 6 and \(6\sqrt{2}\)? No, that's not isosceles. Wait, maybe the right angle is at \((-2, -4)\) or \((4, -4)\). Let's check \((-2, 0)\): distance from \((-2, -4)\) is 4, no. Wait, the two points are \((-2, -4)\) and \((4, -4)\). The vector between them is \((6, 0)\). To form an isosceles right triangle, the other leg should be a vector perpendicular to this, so \((0, 6)\) or \((0, -6)\). So from \((-2, -4)\), adding \((0, 6)\) gives \((-2, 2)\) (not an option), adding \((0, -6)\) gives \((-2, -10)\) (not an option). From \((4, -4)\), adding \((0, 6)\) gives \((4, 2)\) (option B), adding \((0, -6)\) gives \((4, -10)\) (not an option). Wait, but also, the right angle could be at the third point. So the midpoint is \((1, -4)\), and the third point should be such that the distance from \((1, -4)\) to the third point is equal to half the hypotenuse? No, maybe I made a mistake. Wait, let's recalculate the distance between the two given points: \(x\)-coordinates: \(4 - (-2) = 6\), \(y\)-coordinates: \(-4 - (-4) = 0\), so distance is 6. So the legs should be 6 each (since it's isosceles right). So the third point should be 6 units above or below one of the points, vertically, or 6 units horizontally and vertically? Wait, no, in an isosceles right triangle, the legs are equal, so if one leg is horizontal (length 6), the other leg should be vertical (length 6) from one of the points. So from \((-2, -4)\), moving up 6: \(y = -4 + 6 = 2\), so \((-2, 2)\) (not an option). From \((4, -4)\), moving up 6: \(y = -4 + 6 = 2\), so \((4, 2)\) (option B). Let's check the distance from \((-2, -4)\) to \((4, 2)\): \(\sqrt{(4 - (-2))^2 + (2 - (-4))^2}=\sqrt{36 + 36}=\sqrt{72}=6\sqrt{2}\), which is the hypotenuse, and the legs are 6 (horizontal) and 6 (vertical from \((4, -4)\) to \((4, 2)\)). So that's an isosceles right triangle with legs 6 and 6, hypotenuse \(6\sqrt{2}\). Yes, that works. Wait, but earlier I thought the leg was 6 and \(6\sqrt{2}\), but no, the two legs are 6 (horizontal) and 6 (vertical), so it's isosceles right. Let's check other options:
- \((-2, -9)\): distance from \((-2, -4)\) is 5, from \((4, -4)\) is \(\sqrt{6^2 + 5^2}=\sqrt{61}\), not equal.
- \((-2, 0)\): distance from \((-2, -4)\) is 4, from \((4, -4)\) is \(\sqrt{6^2 + 4^2}=\sqrt{52}\), not equal.
- \((3, 1)\): distance from \((-2, -4)\) is \(\sqrt{5^2 + 5^2}=\sqrt{50}\), from \((4, -4)\) is \(\sqrt{1^2 + 5^2}=\sqrt{26}\), not equal.

So the correct point is \((4, 2)\).

Answer:

\((4, 2)\) (Option B: \((4, 2)\))