find the distance between point p and line ℓ.
1. line ℓ contains points (0, -3) and (7, 4). point p has coordinates (4, 3).
2. line ℓ contains points (11, -1) and (-3, -11). point p has coordinates (-1, 1).
3. line ℓ contains points (-2, 1) and (4, 1). point p has coordinates (5, 7).
4. line ℓ contains points (4, -1) and (4, 9). point p has coordinates (1, 6).
5. line ℓ contains points (1, 5) and (4, -4). point p has coordinates (-1, 1).
6. line ℓ contains points (-8, 1) and (3, 1). point p has coordinates (-2, 4).
7. design dante is designing a poster for prom using a design program with a coordinate grid. he starts by creating a geometric border. dante wants the text on the poster to be at least 3 inches away from the top left - hand corner of the border. the border contains the points (0, 7) and (7, 14). if dante places the text at (7, 8), is the text at least 3 inches away from the border? if yes, how far away is the text from the border? let every unit represent an inch. round your answer to the nearest hundredth, if needed.
mixed exercises
find the distance from the line to the given point.
18. y = -3; (5, 2)
19. y = 1/6x + 6; (-6, 5)
20. x = 4; (-2, 5)

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Accepted Answer

1. **For the line $l$ with points $(0, - 3)$ and $(7,4)$ and point $P(4,3)$**:
   - **Step 1: Find the slope of the line $l$**
     - The slope $m$ of a line passing through two - points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $m=\frac{y_2 - y_1}{x_2 - x_1}$. Here, $x_1 = 0,y_1=-3,x_2 = 7,y_2 = 4$, so $m=\frac{4-(-3)}{7 - 0}=\frac{7}{7}=1$.
   - **Step 2: Find the equation of the line $l$ in the point - slope form $y - y_1=m(x - x_1)$**
     - Using the point $(0,-3)$ and $m = 1$, we get $y-(-3)=1(x - 0)$, which simplifies to $y=x - 3$ or $x - y-3=0$.
   - **Step 3: Use the distance formula $d=\frac{\vert Ax_0+By_0 + C\vert}{\sqrt{A^2 + B^2}}$ for a point $(x_0,y_0)$ and a line $Ax+By + C = 0$**
     - Here, $A = 1,B=-1,C=-3,x_0 = 4,y_0 = 3$. Then $d=\frac{\vert1\times4+(-1)\times3-3\vert}{\sqrt{1^2+(-1)^2}}=\frac{\vert4 - 3-3\vert}{\sqrt{2}}=\frac{\vert-2\vert}{\sqrt{2}}=\sqrt{2}$.
2. **For the line $l$ with points $(11,-1)$ and $(-3,-11)$ and point $P(-1,1)$**:
   - **Step 1: Find the slope of the line $l$**
     - $m=\frac{-11-(-1)}{-3 - 11}=\frac{-10}{-14}=\frac{5}{7}$.
   - **Step 2: Find the equation of the line $l$ in the point - slope form using the point $(11,-1)$**
     - $y-(-1)=\frac{5}{7}(x - 11)$, $7(y + 1)=5(x - 11)$, $7y+7 = 5x-55$, $5x-7y-62 = 0$.
   - **Step 3: Use the distance formula**
     - Here, $A = 5,B=-7,C=-62,x_0=-1,y_0 = 1$. Then $d=\frac{\vert5\times(-1)+(-7)\times1-62\vert}{\sqrt{5^2+(-7)^2}}=\frac{\vert-5-7 - 62\vert}{\sqrt{25 + 49}}=\frac{\vert-74\vert}{\sqrt{74}}=\sqrt{74}$.
3. **For the line $l$ with points $(-2,1)$ and $(4,1)$ and point $P(5,7)$**:
   - **Step 1: Find the slope of the line $l$**
     - $m=\frac{1 - 1}{4-(-2)}=0$. The equation of the line is $y = 1$ or $y-1=0$.
   - **Step 2: Use the distance formula**
     - Here, $A = 0,B = 1,C=-1,x_0 = 5,y_0 = 7$. Then $d=\frac{\vert0\times5+1\times7-1\vert}{\sqrt{0^2+1^2}}=\frac{\vert6\vert}{1}=6$.
4. **For the line $l$ with points $(4,-1)$ and $(4,9)$ and point $P(1,6)$**:
   - **Step 1: The line $l$ is a vertical line with $x = 4$ (since $x$ - coordinates of the two points are the same).**
   - **Step 2: The distance between the point $(1,6)$ and the line $x = 4$ is $\vert4 - 1\vert=3$.**
5. **For the line $l$ with points $(1,5)$ and $(4,-4)$ and point $P(-1,1)$**:
   - **Step 1: Find the slope of the line $l$**
     - $m=\frac{-4 - 5}{4 - 1}=\frac{-9}{3}=-3$.
   - **Step 2: Find the equation of the line $l$ using the point - slope form with the point $(1,5)$**
     - $y - 5=-3(x - 1)$, $y-5=-3x + 3$, $3x+y-8 = 0$.
   - **Step 3: Use the distance formula**
     - Here, $A = 3,B = 1,C=-8,x_0=-1,y_0 = 1$. Then $d=\frac{\vert3\times(-1)+1\times1-8\vert}{\sqrt{3^2+1^2}}=\frac{\vert-3 + 1-8\vert}{\sqrt{9 + 1}}=\frac{\vert-10\vert}{\sqrt{10}}=\sqrt{10}$.
6. **For the line $l$ with points $(-8,1)$ and $(3,1)$ and point $P(-2,4)$**:
   - **Step 1: The slope of the line $l$ is $m=\frac{1 - 1}{3-(-8)}=0$, and the equation of the line is $y = 1$ or $y-1=0$.**
   - **Step 2: Use the distance formula**
     - Here, $A = 0,B = 1,C=-1,x_0=-2,y_0 = 4$. Then $d=\frac{\vert0\times(-2)+1\times4-1\vert}{\sqrt{0^2+1^2}}=3$.
7. **For the line $l$ with points $(0,7)$ and $(7,14)$ and point $(7,8)$**:
   - **Step 1: Find the slope of the line $l$**
     - $m=\frac{14 - 7}{7 - 0}=1$. Using the point $(0,7)$, the equation of the line is $y=x + 7$ or $x - y+7=0$.
   - **Step 2: Use the distance formula**
     - Here, $A = 1,B=-1,C = 7,x_0 = 7,y_0 = 8$. Then $d=\frac{\vert1\times7+(-1)\times8+7\vert}{\sqrt{1^2+(-1)^2}}=\frac{\vert7-8 + 7\vert}{\sqrt{2}}=\frac{\vert6\vert}{\sqrt{2}}=3\sqrt{2}\approx4.24$. Since $4.24>3$, the text is at least 3 inches away from the border, and the distance is approximately $4.24$ inches.
18. **For the line $y=-3$ and point $(5,2)$**:
   - The line $y=-3$ is a horizontal line. The distance between the point $(5,2)$ and the line $y=-3$ is $\vert2-(-3)\vert = 5$.
19. **For the line $y=\frac{1}{6}x + 6$ or $x-6y+36 = 0$ and point $(-6,5)$**:
   - Using the distance formula with $A = 1,B=-6,C = 36,x_0=-6,y_0 = 5$, we have $d=\frac{\vert1\times(-6)+(-6)\times5+36\vert}{\sqrt{1^2+(-6)^2}}=\frac{\vert-6-30 + 36\vert}{\sqrt{1 + 36}}=0$.
20. **For the line $x = 4$ and point $(-2,5)$**:
   - The line $x = 4$ is a vertical line. The distance between the point $(-2,5)$ and the line $x = 4$ is $\vert4-(-2)\vert=6$.

# Answer:
1. $\sqrt{2}$
2. $\sqrt{74}$
3. $6$
4. $3$
5. $\sqrt{10}$
6. $3$
7. Yes, approximately $4.24$ inches
18. $5$
19. $0$
20. $6$

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