19) \( A B C \) uchburchak yuzasi 60 ga teng. \( \begin{array}{l}\overrightarrow{B E}=-\frac{1}{2} \overrightarrow{B A} \text { bo'lsa, } \\ A C E \text { uchburchak yuzini toping. }\end{array} \)

Uchburchak ABC ning yuzasi 60 ga teng. Bizga berilgan: \[ \overrightarrow{BE} = -\frac{1}{2} \overrightarrow{BA} \] Bu yerda E nuqtasi BA bo‘ylab B nuqtasi tomonida joylashgan va BA vektorining yarmi bilan teskari yo‘nalishda joylashgan. **Qadamlar:** 1. **Koordinatalarni Belgilash:** - B nuqtasini koordinatalar tizimining boshlang‘ich nuqtasi sifatida qabul qilamiz: \( B(0, 0) \). - \( \overrightarrow{BA} = \mathbf{a} \) vektorini x o‘qi bo‘yicha yo‘lni olgan deb tasavvur qilamiz: \( A(a, 0) \). - \( \overrightarrow{BC} = \mathbf{c} \) vektorini y o‘qi bo‘yicha yo‘lni olgan deb qabul qilamiz: \( C(0, b) \). 2. **Nuqta E ning Koordinatalarini Topish:** \[ \overrightarrow{BE} = -\frac{1}{2} \overrightarrow{BA} = -\frac{1}{2} \mathbf{a} \implies E\left(-\frac{a}{2}, 0\right) \] 3. **ABC Uchburchagini Qayta Ko‘rib Chiqish:** - ABC uchburchagining yuzi: \[ S_{ABC} = \frac{1}{2} \times |a \times b| = 60 \implies a \times b = 120 \] 4. **ACE Uchburchagi Yuzasini Hisoblash:** - A, C, E nuqtalarining koordinatalari: \[ A(a, 0), \quad C(0, b), \quad E\left(-\frac{a}{2}, 0\right) \] - ACE uchburchagining yuzi: \[ S_{ACE} = \frac{1}{2} \times | \vec{AC} \times \vec{AE} | \] \[ \vec{AC} = \left( -a, b \right), \quad \vec{AE} = \left( -\frac{3a}{2}, 0 \right) \] \[ \vec{AC} \times \vec{AE} = (-a) \times 0 - b \times \left(-\frac{3a}{2}\right) = \frac{3ab}{2} \] \[ S_{ACE} = \frac{1}{2} \times \frac{3ab}{2} = \frac{3ab}{4} \] \[ \text{Bizda } ab = 120 \text{ bo‘lgani uchun: } \] \[ S_{ACE} = \frac{3 \times 120}{4} = 90 \] **Javob:** ACE uchburchagining yuzi **90 ga teng**.