1,P is the point (7,5) and L1 is the line with equation 3x+4y=16(1),Find out the equation of the line L2 which passes through P and is perpendicular to L2.(2),Find the point of intersection of the L1 and L2.(3),Find the perpendicular distance of P fr

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1,P is the point (7,5) and L1 is the line with equation 3x+4y=16(1),Find out the equation of the line L2 which passes through P and is perpendicular to L2.(2),Find the point of intersection of the L1 and L2.(3),Find the perpendicular distance of P fr
先翻成中文
P(7,5) 直线L1方程为 3x+4y=16
1.直线过P点且与L1垂直求该直线方程
2.求L1与L2的交点
3.求P点到直线L1的距离
解1.L1:3x+4y=16
y=-3x/4+4
斜率为 -3/4
因为与 L1垂直则L2的斜率为4/3
所以设L2方程为y=4/3x+b
p点代入 b=-13/3
所以L2直线方程为 y=4x/3-13/3
2.联立方程组
y=-3x/4+4
y=4x/3-13/3
解出 x=4,y=1
所以交点坐标为(4,1)
3.点(7,5)到直线距离运用公式
距离=|3x1+4y1-16|/(√(3^2+4^2)=|3*7+5*4-16|/5=5
gradient =y-step/x-step=1/-m
这个方程的意思就是斜率=(y1-y2)/(x1-x2)=1/(-m)
要证明形成的角是直角
只要斜率乘积=-1

直角就是直角,需要理由吗

不是我不想解决，而是你的问题不清楚

翻译一下
P是坐标为(7，5)的点。直线L1方程为3x+4y=16
(1),求过点P且垂直L1的直线L2的方程。
(2),求L1和L2的交点。
(3),求点P到L1的距离。
答案
1.4x-3y=13
2.(4,1)
3. 5
过程
1.先求直线1的斜率，y=-3/4+4,斜率为x的系数，-3/4
两线垂直，...

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翻译一下
P是坐标为(7，5)的点。直线L1方程为3x+4y=16
(1),求过点P且垂直L1的直线L2的方程。
(2),求L1和L2的交点。
(3),求点P到L1的距离。
答案
1.4x-3y=13
2.(4,1)
3. 5
过程
1.先求直线1的斜率，y=-3/4+4,斜率为x的系数，-3/4
两线垂直，则斜率之积为-1，可知线2的斜率为4/3
过(7,5),设方程为y=4/3x+b,代入x=7,y=5,得b=-13/3
整理可得线2方程为4x-3y=13
2.线1和线2联立，解出x,y的值，即为交点坐标
3.点到直线的距离公式：若点为A(m,n),线为Ax+By+C=0
则A到线的距离为｜Am+Bn+C｜/√￣（A^2+B^2）
代入即得答案为5

收起

昏~ 看的不咋明白~
就求出3个点间的距离~满足钩股定理的就是直角
D^2=(X1-X2)^2+(Y1-Y2)^2

这是什么意思#24 汗~~

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