QuestionAugust 26, 2025

Prove that M(-3,2),A(-2,6),I(2,1) and H(1,3) a rhombus. Do we need to use slope or distance formula? __ AT=underline ( )TH=underline ( )HM=underline ( ) Why is this a rhombus? Explain how you know. The solution is below . Did you get it correct : or incorrect? Solution: Rhombus:Parallelogram with 4 congruent sides You only need to use the distance formula to show that all four sides are congruent Any quadrilateral with 4 congruent sides will be a rhombus (and will have parallel sides, so we do not need to prove that part). MA=sqrt (17) MA = AT=sqrt (17) TH=sqrt (17) HM=sqrt (17) HM = All sides are congruent so this is a rhombus

Prove that M(-3,2),A(-2,6),I(2,1) and H(1,3) a rhombus. Do we need to use slope or distance formula? __ AT=underline ( )TH=underline ( )HM=underline ( ) Why is this a rhombus? Explain how you know. The solution is below . Did you get it correct : or incorrect? Solution: Rhombus:Parallelogram with 4 congruent sides You only need to use the distance formula to show that all four sides are congruent Any quadrilateral with 4 congruent sides will be a rhombus (and will have parallel sides, so we do not need to prove that part). MA=sqrt (17) MA = AT=sqrt (17) TH=sqrt (17) HM=sqrt (17) HM = All sides are congruent so this is a rhombus
Prove that M(-3,2),A(-2,6),I(2,1) and H(1,3) a rhombus. Do we need to use slope or distance formula? __ AT=underline ( )TH=underline ( )HM=underline ( ) Why is this a rhombus? Explain how you know. The solution is below . Did you get it correct : or incorrect? Solution: Rhombus:Parallelogram with 4 congruent sides You only need to use the distance formula to show that all four sides are congruent Any quadrilateral with 4 congruent sides will be a rhombus (and will have parallel sides, so we do not need to prove that part). MA=sqrt (17) MA = AT=sqrt (17) TH=sqrt (17) HM=sqrt (17) HM = All sides are congruent so this is a rhombus

Solution
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Answer

Incorrect; the solution provided does not correctly prove that the quadrilateral is a rhombus. Explanation 1. Calculate the distance MA Use the **distance formula**: d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. For M(-3, 2) and A(-2, 6), MA = \sqrt{((-2) - (-3))^2 + (6 - 2)^2} = \sqrt{1^2 + 4^2} = \sqrt{17}. 2. Calculate the distance AT For A(-2, 6) and I(2, 1), AT = \sqrt{(2 - (-2))^2 + (1 - 6)^2} = \sqrt{4^2 + (-5)^2} = \sqrt{16 + 25} = \sqrt{41}. 3. Calculate the distance TH For I(2, 1) and H(1, 3), TH = \sqrt{(1 - 2)^2 + (3 - 1)^2} = \sqrt{(-1)^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}. 4. Calculate the distance HM For H(1, 3) and M(-3, 2), HM = \sqrt{((-3) - 1)^2 + (2 - 3)^2} = \sqrt{(-4)^2 + (-1)^2} = \sqrt{16 + 1} = \sqrt{17}. 5. Verify congruency of all sides Since MA = \sqrt{17}, AT = \sqrt{41}, TH = \sqrt{5}, and HM = \sqrt{17}, not all sides are equal. Therefore, it is not a rhombus.

Explanation

1. Calculate the distance $MA$<br /> Use the **distance formula**: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. For $M(-3, 2)$ and $A(-2, 6)$, $MA = \sqrt{((-2) - (-3))^2 + (6 - 2)^2} = \sqrt{1^2 + 4^2} = \sqrt{17}$.<br /><br />2. Calculate the distance $AT$<br /> For $A(-2, 6)$ and $I(2, 1)$, $AT = \sqrt{(2 - (-2))^2 + (1 - 6)^2} = \sqrt{4^2 + (-5)^2} = \sqrt{16 + 25} = \sqrt{41}$.<br /><br />3. Calculate the distance $TH$<br /> For $I(2, 1)$ and $H(1, 3)$, $TH = \sqrt{(1 - 2)^2 + (3 - 1)^2} = \sqrt{(-1)^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}$.<br /><br />4. Calculate the distance $HM$<br /> For $H(1, 3)$ and $M(-3, 2)$, $HM = \sqrt{((-3) - 1)^2 + (2 - 3)^2} = \sqrt{(-4)^2 + (-1)^2} = \sqrt{16 + 1} = \sqrt{17}$.<br /><br />5. Verify congruency of all sides<br /> Since $MA = \sqrt{17}$, $AT = \sqrt{41}$, $TH = \sqrt{5}$, and $HM = \sqrt{17}$, not all sides are equal. Therefore, it is not a rhombus.
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