Distance of a line from a given point

Question 1: Analyzing a Linear Equation (from file image_015799.png)

The Question: If the general form of a line is $3x + 2y - 5 = 0$, then choose the correct set of options. (Multiple Select Question)

• The slope of the given line is $-\frac{3}{2}$
• The x-intercept is 3
• The point where the given line cuts the X-axis is $(\frac{5}{3}, 0)$
• The y-intercept is 2
• The point where the given line cuts the Y-axis is $(0, \frac{5}{2})$

Core Concepts: Finding Properties from the General Form ($Ax+By+C=0$)

1. Slope ($m$): Rearrange the equation into the slope-intercept form, $y = mx + c$. The coefficient of $x$ is the slope.
2. X-Intercept: This is the point where the line crosses the x-axis, so the y-coordinate is 0. Set $y=0$ and solve for $x$.
3. Y-Intercept: This is the point where the line crosses the y-axis, so the x-coordinate is 0. Set $x=0$ and solve for $y$.

Detailed Solution:

Let’s analyze the equation $3x + 2y - 5 = 0$.

• To find the slope:

  • Rearrange for $y$:
  • $2y = -3x + 5$
  • $y = -\frac{3}{2}x + \frac{5}{2}$
  • The slope ($m$) is the coefficient of $x$. So, the slope is -3/2. The first option is TRUE.

• To find the x-intercept:

  • Set $y=0$:
  • $3x + 2(0) - 5 = 0$
  • $3x - 5 = 0$
  • $3x = 5 \implies x = \frac{5}{3}$.
  • The x-intercept value is 5/3. So, the option “The x-intercept is 3” is FALSE.
  • The point where the line cuts the x-axis is indeed $(\frac{5}{3}, 0)$. The third option is TRUE.

• To find the y-intercept:

  • Set $x=0$:
  • $3(0) + 2y - 5 = 0$
  • $2y - 5 = 0$
  • $2y = 5 \implies y = \frac{5}{2}$.
  • The y-intercept value is 5/2. So, the option “The y-intercept is 2” is FALSE.
  • The point where the line cuts the y-axis is indeed $(0, \frac{5}{2})$. The fifth option is TRUE.

Final Answer: The correct statements are:

• The slope of the given line is $-\frac{3}{2}$
• The point where the given line cuts the X-axis is $(\frac{5}{3}, 0)$
• The point where the given line cuts the Y-axis is $(0, \frac{5}{2})$

Question 2: Distance from a Point to a Line (from file image_015799.png)

The Question: Given the point $(-2, 1)$ and the line $-3x + 4y - 7 = 0$, choose the correct set of options. (Multiple Select Question)

• The distance between $(-2, 1)$ and the line is 2
• The slope of the given line is 5
• The distance between $(-2, 1)$ and the line is $\frac{3}{5}$
• The y-intercept of the given line is $\frac{7}{4}$

Core Concept: Distance from a Point to a Line

The shortest distance ($d$) from a point $(x_1, y_1)$ to a line in the general form $Ax + By + C = 0$ is given by the formula:

Detailed Solution:

Let’s analyze the line $-3x + 4y - 7 = 0$ and the point $(-2, 1)$.

• Slope and y-intercept of the line:

  • Rearrange to $y=mx+c$: $4y = 3x + 7 \implies y = \frac{3}{4}x + \frac{7}{4}$.
  • The slope is 3/4. (So, “The slope…is 5” is FALSE).
  • The y-intercept is 7/4. (So, “The y-intercept…is 7/4” is TRUE).

• Distance from the point to the line:

  • Here, $(x_1, y_1) = (-2, 1)$ and for the line, $A=-3, B=4, C=-7$.
  • Substitute into the distance formula: $$d = \frac{|(-3)(-2) + (4)(1) + (-7)|}{\sqrt{(-3)^2 + 4^2}}$$ $$d = \frac{|6 + 4 - 7|}{\sqrt{9 + 16}}$$ $$d = \frac{|3|}{\sqrt{25}} = \frac{3}{5}$$
  • The distance is 3/5. (So, “The distance…is 2” is FALSE, and “The distance…is 3/5” is TRUE).

Final Answer: The correct options are:

• The distance between $(-2, 1)$ and the line $-3x + 4y - 7 = 0$ is $\frac{3}{5}$
• The y-intercept of the given line is $\frac{7}{4}$

Question 3: Distance Between Parallel Lines (from file image_01575e.png)

The Question: Find the distance between the two parallel lines $4x + 3y + 5 = 0$ and $8x + 6y - 30 = 0$.

Core Concept: Distance Between Parallel Lines

To find the distance between two parallel lines, they must first be written with identical A and B coefficients. For two lines $Ax + By + C_1 = 0$ and $Ax + By + C_2 = 0$, the distance ($d$) is:

Detailed Solution:

1. Line 1: $4x + 3y + 5 = 0$
2. Line 2: $8x + 6y - 30 = 0$
3. Make Coefficients Equal: Notice that the coefficients in the second line are double those in the first. To make them match, divide the entire second equation by 2:
   • $(8x + 6y - 30) \div 2 = 0 \div 2 \implies 4x + 3y - 15 = 0$.
4. Identify Coefficients: Now we compare our two lines:
   • $4x + 3y + 5 = 0 \implies A=4, B=3, C_1=5$
   • $4x + 3y - 15 = 0 \implies A=4, B=3, C_2=-15$
5. Apply the formula: $$d = \frac{|5 - (-15)|}{\sqrt{4^2 + 3^2}}$$ $$d = \frac{|5 + 15|}{\sqrt{16 + 9}}$$ $$d = \frac{|20|}{\sqrt{25}} = \frac{20}{5} = 4$$

Final Answer: The distance between the two parallel lines is 4.

Question 4: Finding ‘c’ from a Given Distance (from file image_01575e.png)

The Question: If the distance between the point $(-1, 3)$ and the line $4x + c = 0$ is $\frac{3}{2}$, then find the value of $c$ where $c$ is a positive number.

Core Concept: Distance from a Point to a Line

We use the same formula as in Question 2: $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.

Detailed Solution:

1. Identify the knowns:
   • The point is $(x_1, y_1) = (-1, 3)$.
   • The line is $4x + c = 0$. We can write this as $4x + 0y + c = 0$.
   • So, $A=4, B=0, C=c$.
   • The distance is given: $d = \frac{3}{2}$.
2. Set up the distance formula equation: $$\frac{3}{2} = \frac{|(4)(-1) + (0)(3) + c|}{\sqrt{4^2 + 0^2}}$$
3. Simplify and solve for c: $$\frac{3}{2} = \frac{|-4 + 0 + c|}{\sqrt{16}}$$ $$\frac{3}{2} = \frac{|c - 4|}{4}$$
4. Isolate the absolute value term:
   • Multiply both sides by 4:
   • $4 \times \frac{3}{2} = |c - 4| \implies 6 = |c - 4|$
5. Solve the absolute value equation:
   • This gives two possibilities:
     • $c - 4 = 6 \implies c = 10$
     • $c - 4 = -6 \implies c = -2$
6. Apply the final condition:
   • The problem states that $c$ is a positive number. Therefore, we discard the $c=-2$ solution.

Final Answer: The value of $c$ is 10.

Question 5: Area of a Triangle (from file image_01575e.png)

The Question: The coordinates of the vertices of a triangle are $(0, 3)$, $(0, 0)$, $(2, 0)$. Find the area of the triangle.

Core Concept: Area of a Right-Angled Triangle

The area of any triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$. When a triangle is on a coordinate plane, if one side lies on the x-axis and another on the y-axis, it is a right-angled triangle, and its base and height are easy to determine.

Detailed Solution:

1. Visualize the vertices:
   • Point A: $(0, 3)$ - on the y-axis.
   • Point B: $(0, 0)$ - the origin.
   • Point C: $(2, 0)$ - on the x-axis.
2. Identify the Base and Height:
   • The segment from the origin (0, 0) to (2, 0) lies on the x-axis. Its length is 2. We can use this as the base.
   • The segment from the origin (0, 0) to (0, 3) lies on the y-axis. Its length is 3. Since it’s perpendicular to the base, we can use this as the height.
3. Calculate the area:
   • Area = $\frac{1}{2} \times \text{base} \times \text{height}$
   • Area = $\frac{1}{2} \times 2 \times 3 = 3$

Final Answer: The area of the triangle is 3 square units.