如L1的点坐标P1(5,9),P2(12,-1), L2的点坐标P1(-4,-8),P2(12,-3),请算出交点坐标.解:将P1(5,9),P2(12,-1),代入L15k+b= 9 12k+b=-1解得:k=-10/7 b=113/7所以直线方程L1:y=(-10/7)x+113/7 (1)式将P1(-4,-8),P2(12,-3)
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![如L1的点坐标P1(5,9),P2(12,-1), L2的点坐标P1(-4,-8),P2(12,-3),请算出交点坐标.解:将P1(5,9),P2(12,-1),代入L15k+b= 9 12k+b=-1解得:k=-10/7 b=113/7所以直线方程L1:y=(-10/7)x+113/7 (1)式将P1(-4,-8),P2(12,-3)](/uploads/image/z/12567979-19-9.jpg?t=%E5%A6%82L1%E7%9A%84%E7%82%B9%E5%9D%90%E6%A0%87P1%285%2C9%29%2CP2%2812%2C-1%29%2C+L2%E7%9A%84%E7%82%B9%E5%9D%90%E6%A0%87P1%28-4%2C-8%29%2CP2%2812%2C-3%29%2C%E8%AF%B7%E7%AE%97%E5%87%BA%E4%BA%A4%E7%82%B9%E5%9D%90%E6%A0%87.%E8%A7%A3%EF%BC%9A%E5%B0%86P1%285%2C9%29%2CP2%2812%2C-1%29%2C%E4%BB%A3%E5%85%A5L15k%2Bb%3D+9+++12k%2Bb%3D-1%E8%A7%A3%E5%BE%97%EF%BC%9Ak%3D-10%2F7++++b%3D113%2F7%E6%89%80%E4%BB%A5%E7%9B%B4%E7%BA%BF%E6%96%B9%E7%A8%8BL1%EF%BC%9Ay%3D%28-10%2F7%29x%2B113%2F7+++++%EF%BC%881%EF%BC%89%E5%BC%8F%E5%B0%86P1%28-4%2C-8%29%2CP2%2812%2C-3%29)
如L1的点坐标P1(5,9),P2(12,-1), L2的点坐标P1(-4,-8),P2(12,-3),请算出交点坐标.解:将P1(5,9),P2(12,-1),代入L15k+b= 9 12k+b=-1解得:k=-10/7 b=113/7所以直线方程L1:y=(-10/7)x+113/7 (1)式将P1(-4,-8),P2(12,-3)
如L1的点坐标P1(5,9),P2(12,-1), L2的点坐标P1(-4,-8),P2(12,-3),请算出交点坐标.
解:将P1(5,9),P2(12,-1),代入L1
5k+b= 9 12k+b=-1
解得:k=-10/7 b=113/7
所以直线方程L1:y=(-10/7)x+113/7
(1)式
将P1(-4,-8),P2(12,-3)代人L2
-4k+b=-8 12k+b=-3 (为什么可以直接代入,数学理论上叫什么?)
解得:k=5/16 b=-27/4
所以直线方程L2:y=(5/16)x-27/4
(2)式
联立上述(1)式,(2)式解方程,结果即是交点坐标
如L1的点坐标P1(5,9),P2(12,-1), L2的点坐标P1(-4,-8),P2(12,-3),请算出交点坐标.解:将P1(5,9),P2(12,-1),代入L15k+b= 9 12k+b=-1解得:k=-10/7 b=113/7所以直线方程L1:y=(-10/7)x+113/7 (1)式将P1(-4,-8),P2(12,-3)
因为点P1,P2都是直线上的点,也就是组成那条直线的一部分,符合直线方程.也没有什么理论,这应该算是常识