Problem 68

Question

Let \(\mathrm{a}, b, c\) and \(d\) be non-zero numbers. If the point of intersection of the lines \(4 a x+2 a y+c=0\) and \(5 b x+2 b y+d\) \(=0\) lies in the fourth quadrant and is equidistant from the two axes then (a) \(3 b c-2 a d=0\) (b) \(3 b c+2 a d=0\) (c) \(2 b c-3 a d=0\) (d) \(2 b c+3 a d=0\)

Step-by-Step Solution

Verified

Answer

Option (a) is correct: \(3bc - 2ad = 0\).

1Step 1: Write the equations of the lines

The two lines are given by the equations \(4ax + 2ay + c = 0\) and \(5bx + 2by + d = 0\). These represent two linear equations in standard form.

2Step 2: Find the point of intersection

To find the point of intersection, solve the system of equations:- Multiply the first equation by \(b\): \(4abx + 2aby + bc = 0\)- Multiply the second equation by \(a\): \(5abx + 2aby + ad = 0\)- Subtract to eliminate \(y\): \((-b c + a d) = 0\)Thus, \(bc = ad\).

3Step 3: Find equidistant point from axes

Since the intersection lies in the fourth quadrant and is equidistant from the x and y axes, the point \((x, y)\) must satisfy \(|x| = |y|\). Thus, we can assume the intersection point \((x, y)\) to be of the form \((k, -k)\), where \(k > 0\).

4Step 4: Substitute point into equations of lines

Substitute \((k, -k)\) into the first equation: \(4ak - 2ak + c = 0\) gives \(2ak = c\).Substitute \((k, -k)\) into the second equation: \(5bk - 2bk + d = 0\) gives \(3bk = d\).

5Step 5: Substitute expressions back into condition

We have \(bc = ad\).Now replace \(c = 2ak\) and \(d = 3bk\) in the condition:- Substituting gives \(b(2ak) = a(3bk)\) or \(2bak = 3abk\).- Cancel \(ak\), and we obtain: \(2b = 3a\).

6Step 6: Check the options

From \(2b = 3a\), if we choose option (a): \(3bc - 2ad = 0\), substitute \(c = 2ak\) and \(d = 3bk\), we have - \(3b(2ak) - 2a(3bk) = 6bak - 6abk = 0\).Thus, option (a) is satisfied.

Key Concepts

Quadrant SystemEquation of a LinePoint of IntersectionDistance from Axes

Quadrant System

The quadrant system is a way to categorize locations on the Cartesian coordinate plane. This plane is divided into four sections, called quadrants, by the x-axis and y-axis. Each quadrant is defined by the signs of the coordinates:

First Quadrant (I): Both x and y are positive.
Second Quadrant (II): x is negative, y is positive.
Third Quadrant (III): Both x and y are negative.
Fourth Quadrant (IV): x is positive, y is negative.

In the context of the exercise, the point of intersection lies in the fourth quadrant, meaning it has coordinates (x, -y) where x > 0 and y > 0. This is important because it helps us understand the nature of the point's position relative to the axes.

Equation of a Line

The equation of a line in the coordinate plane is often expressed in the standard form as Ax + By + C = 0, where A, B, and C are constants. This form allows us to describe a line by its slope and position relative to the coordinate axes.

In our original exercise, the lines given are:

4ax + 2ay + c = 0
5bx + 2by + d = 0

These equations appear in standard form and represent two different straight lines on the Cartesian plane. By understanding how to manipulate and solve these equations, we can find important features, such as intersections and slopes, which are critical for solving geometric problems.

Point of Intersection

The point of intersection of two lines is where they meet, or cross, each other on the Cartesian plane. To find this point, you need to solve the equations of the lines simultaneously. This involves combining the equations to eliminate one of the variables, allowing the calculation of the other.

In the exercise, we perform the following steps:

Multiply the first line equation by b: 4abx + 2aby + bc = 0.
Multiply the second line equation by a: 5abx + 2aby + ad = 0.
Subtract them to eliminate y, resulting in: -bc + ad = 0, or bc = ad.

This process gives us the condition that must be met by the parameters of the lines for them to intersect at a point. Knowing the quadrant also helps refine the possible solutions.

Distance from Axes

In coordinate geometry, the distance from a point to the axes can tell us a lot about its position. If a point is equidistant from the x and y axes, it means the absolute values of its x-coordinate and y-coordinate are equal. This is especially useful when the problem specifies a condition like being equidistant from the axes.

For our exercise, we assume the point of intersection is (k, -k) because it lies in the fourth quadrant and is equidistant from the axes. Hence,