如图1,y=-x+6与坐标轴交于A.B两点,点C在X轴的负半轴上,S△OBC=1/3S△AOB (1)求直线BC的解析式 (2)直线EF:y=kx-k交AB于E点,与X轴交于D点,交BC的延长线于点F,且S△BED=S△FBD,求k的值 (3)如图2,M(2,
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![如图1,y=-x+6与坐标轴交于A.B两点,点C在X轴的负半轴上,S△OBC=1/3S△AOB (1)求直线BC的解析式 (2)直线EF:y=kx-k交AB于E点,与X轴交于D点,交BC的延长线于点F,且S△BED=S△FBD,求k的值 (3)如图2,M(2,](/uploads/image/z/12412634-50-4.jpg?t=%E5%A6%82%E5%9B%BE1%2Cy%3D-x%2B6%E4%B8%8E%E5%9D%90%E6%A0%87%E8%BD%B4%E4%BA%A4%E4%BA%8EA.B%E4%B8%A4%E7%82%B9%2C%E7%82%B9C%E5%9C%A8X%E8%BD%B4%E7%9A%84%E8%B4%9F%E5%8D%8A%E8%BD%B4%E4%B8%8A%2CS%E2%96%B3OBC%3D1%2F3S%E2%96%B3AOB+%EF%BC%881%EF%BC%89%E6%B1%82%E7%9B%B4%E7%BA%BFBC%E7%9A%84%E8%A7%A3%E6%9E%90%E5%BC%8F+%EF%BC%882%EF%BC%89%E7%9B%B4%E7%BA%BFEF%EF%BC%9Ay%3Dkx-k%E4%BA%A4AB%E4%BA%8EE%E7%82%B9%2C%E4%B8%8EX%E8%BD%B4%E4%BA%A4%E4%BA%8ED%E7%82%B9%2C%E4%BA%A4BC%E7%9A%84%E5%BB%B6%E9%95%BF%E7%BA%BF%E4%BA%8E%E7%82%B9F%2C%E4%B8%94S%E2%96%B3BED%3DS%E2%96%B3FBD%2C%E6%B1%82k%E7%9A%84%E5%80%BC+%EF%BC%883%EF%BC%89%E5%A6%82%E5%9B%BE2%2CM%EF%BC%882%2C)
如图1,y=-x+6与坐标轴交于A.B两点,点C在X轴的负半轴上,S△OBC=1/3S△AOB (1)求直线BC的解析式 (2)直线EF:y=kx-k交AB于E点,与X轴交于D点,交BC的延长线于点F,且S△BED=S△FBD,求k的值 (3)如图2,M(2,
如图1,y=-x+6与坐标轴交于A.B两点,点C在X轴的负半轴上,S△OBC=1/3S△AOB (1)求直线BC的解析式 (2)直线EF:y=kx-k交AB于E点,与X轴交于D点,交BC的延长线于点F,且S△BED=S△FBD,求k的值 (3)如图2,M(2,4),点P为x轴上一动点,AH垂直PM垂足为H,取HG=HA,连CG,当P点运动时,∠CGM大小是否变化,并给予证明.
如图1,y=-x+6与坐标轴交于A.B两点,点C在X轴的负半轴上,S△OBC=1/3S△AOB (1)求直线BC的解析式 (2)直线EF:y=kx-k交AB于E点,与X轴交于D点,交BC的延长线于点F,且S△BED=S△FBD,求k的值 (3)如图2,M(2,
(l)直线y=一x十6,A(6,o),B(0,6)又因为s△B0c=s△A0B/3所以c(一2,0)直线Bc斜率K=6一0/0+2=3所以Bc方程是:y=3X十6