如图,直线y=-1/2x+2交x轴于A点,交y轴于B点,点P为双曲线y+k/x(x>0)上一点,且PA=PB,如图,直线y=-1/2x+2交x轴于A点,交y轴于B点,点P为双曲线y+k/x(x>0)上一点,∠APB=90°,求k的值.
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![如图,直线y=-1/2x+2交x轴于A点,交y轴于B点,点P为双曲线y+k/x(x>0)上一点,且PA=PB,如图,直线y=-1/2x+2交x轴于A点,交y轴于B点,点P为双曲线y+k/x(x>0)上一点,∠APB=90°,求k的值.](/uploads/image/z/8654091-51-1.jpg?t=%E5%A6%82%E5%9B%BE%2C%E7%9B%B4%E7%BA%BFy%3D-1%2F2x%2B2%E4%BA%A4x%E8%BD%B4%E4%BA%8EA%E7%82%B9%2C%E4%BA%A4y%E8%BD%B4%E4%BA%8EB%E7%82%B9%2C%E7%82%B9P%E4%B8%BA%E5%8F%8C%E6%9B%B2%E7%BA%BFy%2Bk%2Fx%28x%3E0%29%E4%B8%8A%E4%B8%80%E7%82%B9%2C%E4%B8%94PA%3DPB%2C%E5%A6%82%E5%9B%BE%2C%E7%9B%B4%E7%BA%BFy%3D-1%2F2x%2B2%E4%BA%A4x%E8%BD%B4%E4%BA%8EA%E7%82%B9%2C%E4%BA%A4y%E8%BD%B4%E4%BA%8EB%E7%82%B9%2C%E7%82%B9P%E4%B8%BA%E5%8F%8C%E6%9B%B2%E7%BA%BFy%2Bk%2Fx%28x%3E0%29%E4%B8%8A%E4%B8%80%E7%82%B9%2C%E2%88%A0APB%3D90%C2%B0%2C%E6%B1%82k%E7%9A%84%E5%80%BC.)
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如图,直线y=-1/2x+2交x轴于A点,交y轴于B点,点P为双曲线y+k/x(x>0)上一点,且PA=PB,如图,直线y=-1/2x+2交x轴于A点,交y轴于B点,点P为双曲线y+k/x(x>0)上一点,∠APB=90°,求k的值.
如图,直线y=-1/2x+2交x轴于A点,交y轴于B点,点P为双曲线y+k/x(x>0)上一点,且PA=PB,
如图,直线y=-1/2x+2交x轴于A点,交y轴于B点,点P为双曲线y+k/x(x>0)上一点,∠APB=90°,求k的值.
如图,直线y=-1/2x+2交x轴于A点,交y轴于B点,点P为双曲线y+k/x(x>0)上一点,且PA=PB,如图,直线y=-1/2x+2交x轴于A点,交y轴于B点,点P为双曲线y+k/x(x>0)上一点,∠APB=90°,求k的值.
直线CD:y=-1/2x+2与x轴交于C(4,0)与y轴交于D(0,2),AD方程:y=2x+2,得A点坐标为(-1,0),AB方程:y=-1/2x-1/2,BC方程:y=2x-8,由AB与BC组成方程组得B(3,-2)因为双曲线y=k/x(k<0)经过点B,所以-2=k/3,得k=-6,又由y=-1/2x+2和y=-6/x联立解得:E(6,-1),M(6,0)作BG⊥x轴于G,则G(3,0)S BEMC=S梯形BEMG-S三角形BCG=(1+2)*3/2-(4-3)*2/2=7/2