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7. (内江中考)在平面直角坐标系中,横坐标和纵坐标都是整数的点叫作整点,已知直线$y = tx + 2t + 2(t＞0)$与两坐标轴围成的三角形区域(不含边界)中有且只有四个整点,则$t$的取值范围是()
A. $\frac{1}{2}≤t＜2$
B. $\frac{1}{2}＜t≤1$
C. $1＜t≤2$
D. $\frac{1}{2}≤t≤2且t≠1$
A. $\frac{1}{2}≤t＜2$
B. $\frac{1}{2}＜t≤1$
C. $1＜t≤2$
D. $\frac{1}{2}≤t≤2且t≠1$
答案
D 解析:$\because y=tx+2t+2=t(x+2)+2(t＞0),\therefore $直线$y=tx+2t+2(t＞0)$经过点$(-2,2)$.如图,当直线经过点$(0,3)$时,直线$y=tx+2t+2(t＞0)$与两坐标轴围成的三角形区域(不含边界)中有且只有四个整点,则$3=2t+2$,解得$t=\frac {1}{2}$;当直线经过点$(0,6)$时,直线$y=tx+2t+2(t＞0)$与两坐标轴围成的三角形区域(不含边界)中有且只有四个整点,则$6=2t+2$,解得$t=2$;当直线经过点$(0,4)$时,直线$y=tx+2t+2(t＞0)$与两坐标轴围成的三角形区域(不含边界)中有且只有三个整点,则$4=2t+2$,解得$t=1;\therefore $直线$y=tx+2t+2(t＞0)$与两坐标轴围成的三角形区域(不含边界)中有且只有四个整点,则t的取值范围是$\frac {1}{2}≤t≤2$且$t≠1$.
8. 已知点$A(2,-3)$,$B(-3,-2)$,直线$l:y = -kx + k + 1与线段AB$相交,则$k$的取值范围是______.
答案
$k≤-\frac {3}{4}$或$k≥4$ 解析:如图,直线$l:y=-kx+k+1=k(1-x)+1$经过定点$C(1,1)$,当直线l经过点$B(-3,-2)$时,直线BC的表达式为$y=\frac {3}{4}x+\frac {1}{4}$,结合图形知$-k≥\frac {3}{4},\therefore k≤-\frac {3}{4}$;当直线l经过点$A(2,-3)$时,直线AC的表达式为$y=-4x+5$,结合图形知$-k≤-4,\therefore k≥4$.综上可知,$k≤-\frac {3}{4}$或$k≥4$.
9. (2024·武汉期中)小明用几何画板作出一函数图象,若
有点$P(m^2,\frac{m^2}{2} - \frac{1}{2})$在该函数图象上,则这样的点$P$有______个.
有点$P(m^2,\frac{m^2}{2} - \frac{1}{2})$在该函数图象上,则这样的点$P$有______个.
答案
2 解析:$\because m^{2}≥0$,将点P看作是直线$y=\frac {1}{2}k-\frac {1}{2}(k=m^{2}≥0)$上的点,令$m^{2}=0$,则$y=-\frac {1}{2}$,令$m^{2}=1$,则$y=0$,则直线$y=\frac {1}{2}k-\frac {1}{2}$$(k=m^{2}≥0)$过点$(0,-\frac {1}{2}),(1,0)$,如图,作出函数图象,则点$P(m^{2},\frac {m^{2}}{2}-\frac {1}{2})$在该函数图象上有2个.
10. 已知平面直角坐标系中,点$P(x_0,y_0)和直线Ax + By + C = 0$(其中$A$,$B$不全为0),则点$P到直线Ax + By + C = 0的距离d可用公式d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$来计算.
例如：求点$P(0,0)到直线4x + 3y - 3 = 0$的距离,因为$A = 4$,$B = 3$,$C = -3$,所以距离$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} = \frac{|4×0 + 3×0 - 3|}{\sqrt{4^2 + 3^2}} = \frac{3}{5}$.
(1)点$M(3,4)到直线y = -\frac{3}{4}x + \frac{5}{4}$的距离为______；
(2)如图,直线$y = -x沿y$轴向上平移1个单位长度得到另一条直线,求这两条平行直线之间的距离；
(3)若点$M(1,0)到直线x + y + C = 0的距离为\sqrt{2}$,求实数$C$的值.
例如：求点$P(0,0)到直线4x + 3y - 3 = 0$的距离,因为$A = 4$,$B = 3$,$C = -3$,所以距离$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} = \frac{|4×0 + 3×0 - 3|}{\sqrt{4^2 + 3^2}} = \frac{3}{5}$.
(1)点$M(3,4)到直线y = -\frac{3}{4}x + \frac{5}{4}$的距离为______；
(2)如图,直线$y = -x沿y$轴向上平移1个单位长度得到另一条直线,求这两条平行直线之间的距离；
(3)若点$M(1,0)到直线x + y + C = 0的距离为\sqrt{2}$,求实数$C$的值.
答案
(1)4 解析:$\because y=-\frac {3}{4}x+\frac {5}{4}$可变形为$3x+4y-5=0$,则其中$A=3,$$B=4,C=-5$,由公式得,$d=\frac {|9+16-5|}{\sqrt {3^{2}+4^{2}}}=\frac {20}{5}=4,\therefore $点$M(3,4)$到直线$y=-\frac {3}{4}x+\frac {5}{4}$的距离为4.
(2)平移后的直线表达式为$y=-x+1$,在直线$y=-x$上任意取一点P,当$x=0$时,$y=0.\therefore P(0,0).\because $直线$y=-x+1,\therefore x+y-1=0,A=1,$$B=1,C=-1,\therefore d=\frac {|0+0-1|}{\sqrt {1^{2}+1^{2}}}=\frac {1}{\sqrt {2}}=\frac {\sqrt {2}}{(\sqrt {2})^{2}}=\frac {\sqrt {2}}{2}.\therefore $两平行线之间的距离为$\frac {\sqrt {2}}{2}$. (注:八下将会学习二次根式的运算,$\frac {1}{\sqrt {2}}=\frac {\sqrt {2}}{(\sqrt {2})^{2}}=\frac {\sqrt {2}}{2})$
(3)由点到直线的距离公式得$\frac {|1×1+1×0+C|}{\sqrt {1^{2}+1^{2}}}=\frac {|1+C|}{\sqrt {2}}=\sqrt {2},$$\therefore |1+C|=(\sqrt {2})^{2}=2$,解得$C=1$或$C=-3$.
(2)平移后的直线表达式为$y=-x+1$,在直线$y=-x$上任意取一点P,当$x=0$时,$y=0.\therefore P(0,0).\because $直线$y=-x+1,\therefore x+y-1=0,A=1,$$B=1,C=-1,\therefore d=\frac {|0+0-1|}{\sqrt {1^{2}+1^{2}}}=\frac {1}{\sqrt {2}}=\frac {\sqrt {2}}{(\sqrt {2})^{2}}=\frac {\sqrt {2}}{2}.\therefore $两平行线之间的距离为$\frac {\sqrt {2}}{2}$. (注:八下将会学习二次根式的运算,$\frac {1}{\sqrt {2}}=\frac {\sqrt {2}}{(\sqrt {2})^{2}}=\frac {\sqrt {2}}{2})$
(3)由点到直线的距离公式得$\frac {|1×1+1×0+C|}{\sqrt {1^{2}+1^{2}}}=\frac {|1+C|}{\sqrt {2}}=\sqrt {2},$$\therefore |1+C|=(\sqrt {2})^{2}=2$,解得$C=1$或$C=-3$.
