1,P is the point (7,5) and L1 is the line with equation 3x+4
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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 from the line L1.
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可怜小妹的数学初中水平还不够,看来勘定英文是不够的!555555555555555555555555
还有一个公式:gradient =y-step/x-step=1/-m
给出3个点的坐标,如何证明他们形成的角是直角?(use m1m2=-1 to explain it)
(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 from the line L1.
------------------------------------------------
可怜小妹的数学初中水平还不够,看来勘定英文是不够的!555555555555555555555555
还有一个公式:gradient =y-step/x-step=1/-m
给出3个点的坐标,如何证明他们形成的角是直角?(use m1m2=-1 to explain it)
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
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
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