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A quadrilateral has its vertices at points A(-4,3), B(-2,3), C(-2,1), and D(-4,1). a) Plot the points on a cartesian plane and from them, what is the shape of the figure formed?

Mathematics

ScanSolve AI Answer

Answer

Mraba\text{Mraba}

Step-by-step solution

5 steps

Huu hapa utatuzi wa swali lako:

a)

  1. Andika nukta zilizopewa.

    Nukta za pembe nne ni:

    A(−4,3)A(-4, 3) B(−2,3)B(-2, 3) C(−2,1)C(-2, 1) D(−4,1)D(-4, 1)
  2. Jinsi ya kuchora nukta kwenye mchoro wa Cartesia.

    Ili kuchora nukta hizi, chora mhimili wa x (mlalo) na mhimili wa y (wima) unaokatana kwenye asili (0,0). Kisha, kwa kila nukta (x,y)(x, y):

    • Anza kwenye asili.
    • Sogea xx vitengo kushoto (kama xx ni hasi) au kulia (kama xx ni chanya).
    • Kutoka hapo, sogea yy vitengo chini (kama yy ni hasi) au juu (kama yy ni chanya).
    • Weka alama kwenye eneo hilo. Baada ya kuchora nukta zote nne, ziunganishe kwa mpangilio (A hadi B, B hadi C, C hadi D, na D hadi A) ili kuunda pembe nne.
  3. Kokotoa urefu wa kila upande.

    Tutatumia fomula ya umbali d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.

    Urefu wa AB:

    AB=(−2−(−4))2+(3−3)2AB = \sqrt{(-2 - (-4))^2 + (3 - 3)^2} AB=(2)2+(0)2AB = \sqrt{(2)^2 + (0)^2} AB=4+0AB = \sqrt{4 + 0} AB=4=2 vitengoAB = \sqrt{4} = 2 \text{ vitengo}

    Urefu wa BC:

    BC=(−2−(−2))2+(1−3)2BC = \sqrt{(-2 - (-2))^2 + (1 - 3)^2} BC=(0)2+(−2)2BC = \sqrt{(0)^2 + (-2)^2} BC=0+4BC = \sqrt{0 + 4} BC=4=2 vitengoBC = \sqrt{4} = 2 \text{ vitengo}

    Urefu wa CD:

    CD=(−4−(−2))2+(1−1)2CD = \sqrt{(-4 - (-2))^2 + (1 - 1)^2} CD=(−2)2+(0)2CD = \sqrt{(-2)^2 + (0)^2} CD=4+0CD = \sqrt{4 + 0} CD=4=2 vitengoCD = \sqrt{4} = 2 \text{ vitengo}

    Urefu wa DA:

    DA=(−4−(−4))2+(3−1)2DA = \sqrt{(-4 - (-4))^2 + (3 - 1)^2} DA=(0)2+(2)2DA = \sqrt{(0)^2 + (2)^2} DA=0+4DA = \sqrt{0 + 4} DA=4=2 vitengoDA = \sqrt{4} = 2 \text{ vitengo}
  4. Kokotoa miteremko ya kila upande.

    Tutatumia fomula ya mteremko m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}.

    Mteremko wa AB:

    mAB=3−3−2−(−4)=02=0m_{AB} = \frac{3 - 3}{-2 - (-4)} = \frac{0}{2} = 0

    (Mstari mlalo)

    Mteremko wa BC:

    mBC=1−3−2−(−2)=−20m_{BC} = \frac{1 - 3}{-2 - (-2)} = \frac{-2}{0}

    (Mteremko haujafafanuliwa, mstari wima)

    Mteremko wa CD:

    mCD=1−1−4−(−2)=0−2=0m_{CD} = \frac{1 - 1}{-4 - (-2)} = \frac{0}{-2} = 0

    (Mstari mlalo)

    Mteremko wa DA:

    mDA=3−1−4−(−4)=20m_{DA} = \frac{3 - 1}{-4 - (-4)} = \frac{2}{0}

    (Mteremko haujafafanuliwa, mstari wima)

  5. Tambua umbo la umbo lililoundwa.

    • Tumegundua kuwa pande zote nne (AB, BC, CD, DA) zina urefu sawa wa vitengo 2.
    • Mstari AB na CD ni milalo (mteremko 0), hivyo ni sambamba.
    • Mstari BC na DA ni wima (mteremko haujafafanuliwa), hivyo ni sambamba.
    • Pande zinazopakana (kama AB na BC) ni perpendicular kwa sababu moja ni mlalo na nyingine ni wima.

    Kwa kuwa pande zote zina urefu sawa na pembe zote ni pembe mraba (digrii 90), umbo lililoundwa ni mraba.

    Umbo la umbo lililoundwa ni: Mraba\text{Mraba}

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