Tambayoyi Misali Game da Matsayin Wani Maudu'i Dangane da Da'ira
Tantance matsayin wuri dangane da da'ira muhimmin batu ne a fannin lissafi na asali, musamman a nazarin da'irori. A cikin wannan labarin, za mu tattauna misalai da dama da suka shafi matsayin wuri dangane da da'ira, tare da bayaninsu. Wannan zai taimaka wajen fayyace manufar ta hanyar amfani da aikace-aikace.
Pendahuluan
Kafin mu shiga cikin tambayoyin misalai, bari mu tuna da matsayi uku na wani wuri a kan da'ira:
1. A cikin da'ira: Idan nisan da'ira zuwa tsakiyar da'irar ya fi ƙanƙanta fiye da radius na da'irar.
2. A wajen da'irar: Idan nisan wurin zuwa tsakiyar da'irar ya fi radius na da'irar girma.
3. A kan da'ira: Idan nisan da ke tsakanin wurin zuwa tsakiyar da'irar yayi daidai da radius na da'irar.
A fannin lissafi, ana iya tantance matsayin wurin \(T(x_1, y_1)\) dangane da da'irar da ke tsakiya a \((a, b)\) tare da radius \(r\) ta hanyar kwatanta \(T(x_1, y_1)\) da lissafin da'irar, wato:
\[
(x – a)^2 + (y – b)^2 = r^2
\]
Idan sakamakon maye gurbin \(x_1\) da \(y_1\) a cikin lissafi ya ba da ƙimar:
– Ƙarami fiye da \(r^2\), ma'anar tana cikin da'irar.
– Fiye da \(r^2\), ma'anar tana wajen da'irar.
– Kamar yadda \(r^2\), ma'anar tana kan da'irar.
Tambayoyi da Tattaunawa Samfura
Tambaya ta 1
Kayyade matsayin maki \(T(3, 4)\) dangane da da'irar da ke da lissafin \( (x – 1)^2 + (y – 2)^2 = 25 \).
Tattaunawa:
Mataki na farko shine a kimanta lissafin da'irar sannan a nemo nisan daga wurin \(T(3, 4)\) zuwa tsakiyar da'irar \((1, 2)\).
1. Gano tsakiyar da kuma radius na da'irar:
Daidaito na da'ira: \( (x – 1)^2 + (y – 2)^2 = 25 \)
– Cibiyar da'ira (\(a, b\)): (1, 2)
– Radius na da'irar (\(r\)): \(\sqrt{25} = 5\)
2. Lissafa nisan da ke tsakanin maki \(T(3, 4)\) da tsakiyar da'irar \( (1, 2) \):
\[
D = \sqrt{(3 – 1)^2 + (4 – 2)^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}
\]
Ƙimar \( 2\sqrt{2} \approx 2 \sau 1.414 = 2.828 \) (ƙasa da \(5\)).
3. Kammalawa:
Tunda \( 2\sqrt{2} < 5 \), to ma'anar \( T(3, 4) \) tana cikin da'irar. Tambaya ta 2 Da'ira tana da tsakiya a wurin \( (0, 0) \) da kuma radius na 7. Kayyade matsayin wurin \(P(5, 6)\) dangane da da'irar.
Tambaya ta 3
Kayyade matsayin maki \(M(2, -1)\) dangane da da'irar da ke da lissafin \( x^2 + y^2 = 5 \).
Tattaunawa:
1. Lissafa nisan daga maki \(M(2, -1)\) zuwa tsakiyar da'irar \( (0, 0):
\[
D = \sqrt{(2 – 0)^2 + (-1 – 0)^2} = \sqrt{4 + 1} = \sqrt{5}
\]
2. Kwatanta nisan \(D\) da radius na da'irar:
Radius na da'ira (\(r\)) = \(\sqrt{5}\).
3. Kammalawa:
Tunda \( \sqrt{5} = \sqrt{5} \), to, ma'anar \( M(2, -1) \) tana kan da'irar.
Tambaya ta 4
Da'ira mai tsakiya a \( (4, 3) \) tana da radius \(\sqrt{10}\). Nuna matsayin wurin \(N(7, 7) \) dangane da wannan da'irar.
Tattaunawa:
1. Daidaito na Da'ira:
Daidaiton da'ira mai tsakiya \( (4, 3) \) da radius \( \sqrt{10} \) shine:
\[
(x – 4)^2 + (y – 3)^2 = 10
\]
2. Lissafa nisan daga wurin \( N(7, 7) \) zuwa tsakiyar da'irar \( (4, 3) \):
\[
D = \sqrt{(7 – 4)^2 + (7 – 3)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5
\]
3. Kammalawa:
Tunda \( 5 > \sqrt{10} \), to, ma'anar \( N(7, 7) \) tana wajen da'irar.
Penutup
Ta hanyar fahimtar yadda ake ƙididdige nisan da ke tsakanin wani wuri da tsakiyar da'ira da kuma kwatanta shi da radius, za mu iya tantance matsayin wani wuri dangane da da'irar cikin sauƙi. Ana sa ran tattaunawar da ke cikin wannan labarin za ta samar da fahimtar manufar da kuma yadda za a magance matsalolin da suka shafi matsayin wani wuri dangane da da'ira.
A aikace, sanin matsayin waɗannan batutuwa yana da matuƙar amfani a aikace-aikacen lissafi daban-daban, ciki har da nazarin lissafi, ƙirar zane-zane, da injiniyanci. Saboda haka, ƙwarewa a wannan ra'ayi muhimmin tushe ne wanda ya cancanci kulawa sosai da kuma fahimtar juna.
