Question

Find the equation of a line which divides the line segment joining the points \((1,-2)\) and \((3,-1)\) in the ratio \(3: 1\) perpendicularly. (1) \(x-2 y-5=0\) (2) \(6 x+4 y-5=0\) (3) \(3 x+2 y-5=0\) (4) \(8 x+4 y-15=0\)

Short answer

Question: Find the equation of the line that divides the line segment joining (1, -2) and (3, -1) in the ratio 3:1 perpendicularly. Answer: The equation of the required line is 8x + 4y - 5 = 0.

Step-by-step solution

Coordinates of the dividing point

To find the coordinates of the point dividing the line segment in the ratio \(3:1\), we will use the section formula. Let the point be \((x, y)\), then we can find the coordinates as follows: \(x = \frac{(3 * 1) + (1 * 3)}{3 + 1} = \frac{6}{4} = \frac{3}{2}\) \(y = \frac{(3 * (-2)) + (1 * (-1))}{3 + 1} = \frac{-7}{4}\) The dividing point is \((\frac{3}{2}, \frac{-7}{4})\).

Slope of the line segment

To find the slope of the line segment, we will use the slope formula between the two points \((1, -2)\) and \((3, -1)\): \(m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 + 2}{3 - 1} = \frac{1}{2}\)

Perpendicular slope

Now we need to find the negative reciprocal of the slope, which is the slope of the perpendicular line dividing the line segment: \(m_{\perp} = -\frac{1}{m} = -\frac{1}{\frac{1}{2}} = -2\)

Equation of the line

Now we can find the equation of the line using the point-slope form, which is \(y - y_1 = m_{\perp}(x - x_1)\). Substituting the values we found, we get: \(y - \frac{-7}{4} = -2(x - \frac{3}{2})\) Now let's simplify and put the equation in the form \(ax + by + c = 0\): \(y + \frac{7}{4} = -2x + 3\) \(4y + 7 = -8x + 12\) \(8x + 4y - 5 = 0\) The equation of the required line is \(\boxed{8x + 4y - 5 = 0}\), which corresponds to option (4).

Key concepts

Point-Slope Form

The point-slope form is a method used to find the equation of a line when you know the slope and one point on the line. This form is expressed mathematically as:\[ y - y_1 = m(x - x_1) \]Where \( (x_1, y_1) \) is a known point on the line, and \( m \) is the slope of the line.
Using this form allows us to easily plug in known values to find an equation that describes the entire line.
A real-world application, like in solving our exercise, is when you need a perpendicular line passing through a specific point - here, the midpoint calculated using the section formula.

Section Formula

The section formula helps in finding a point that divides a line segment into a specific ratio. This point's coordinates can be calculated if its division ratio and the endpoint coordinates are known. The formula is:

• For the x-coordinate: \[ x = \frac{mx_2 + nx_1}{m+n} \]
• For the y-coordinate: \[ y = \frac{my_2 + ny_1}{m+n} \]

This method is crucial in our exercise to determine the specific dividing point of the segment joining
points \((1, -2)\) and \((3, -1)\) in the ratio \(3:1\). The calculated coordinates then serve as a reference point in forming the line equation with the point-slope form.

Perpendicular Slope

To find the slope of a line perpendicular to a given line, we take the negative reciprocal of the original line's slope. If the slope of the original line is \( m \), then the perpendicular slope \( m_{\perp} \) is:\[ m_{\perp} = -\frac{1}{m} \]In simpler terms, this means you flip the original slope’s fraction and change its sign.
This concept is important because it helps us find how the new line interacts with the original segment - essential in our exercise for determining the line dividing the given segment perpendicularly.

Line Segment Division

Line segment division involves finding a point that divides the segment either internally or externally. This is often necessary for various geometrical constructions and calculations. When a point divides a line segment in a specific ratio, either calculated using the section formula or based on context, it determines various properties such as length divisions and orientations.
The division point was used in our exercise not only for identifying the segment's dividing point but also for establishing it as the centerpiece of a perpendicular bisector necessary for our line equation.