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CAT 2023 Percentile Predictor

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CAT Area between the lines - Practice Questions & MCQ

Edited By admin | Updated on Oct 04, 2023 04:20 PM | #CAT

Quick Facts

  • 5 Questions around this concept.

Solve by difficulty

A straight line L through the point (3, -2) is inclined at an angle of 60o to the line \sqrt{3}\; x+y=1 . If L also intersects the x-axis, then the equation of L is

 

Line L_{1} is perpendicular to \sqrt{3}x+y= 0. If L_{2} makes an angle of 45^{\circ} with L_{1}, then the slope of L_{2} can be

The lines p\left ( p^{2}+1 \right )x-y+q= 0 and

\left ( p^{2}+1 \right )^{2}\! x+\left ( p^{2}+1 \right )y+2q= 0 are perpendicular to a common line for

Find the slopes of the straight lines which are equally inclined to lines 3x-4y-7= 0 and 12x-5y+6= 0.

Concepts Covered - 1

Area between the lines

Given two linear equations in two variables of the form \mathrm{\( ax + by + c1 = 0 \) and \( ax + by + c2 = 0 \)}, the distance between these two parallel lines can be given as:

\mathrm{\[ d = \frac{{|c1 - c2|}}{{\sqrt{a^2 + b^2}}} \]}

When we talk about the area between two lines, it usually refers to the area bounded by these lines on a graph.

- Before finding the area between two lines, it's essential to ensure they are parallel. If two lines are given by \mathrm{\( ax + by + c1 = 0 \) and \( ax + by + c2 = 0 \)}, they are parallel since their directional ratios are the same.

  - If the lines intersect, the area between them doesn't exist in the traditional sense, but we can consider the area of the shape (like a triangle or quadrilateral) formed by these lines and the x or y axes.

Foundation Building Questions:

Question:

Given the equations:

1) \mathrm{\( 3x + 4y - 12 = 0 \)}

2) \mathrm{ \( 3x + 4y + 15 = 0 \)}

Find the area between these two lines up to the x-axis.

Solution:

First, ensure the lines are parallel. The coefficients of (x) and (y) are the same, so they are parallel.

To find the points where these lines intersect the x-axis, set \mathrm{ \( y = 0 \)} in both equations:

1) \mathrm{\( 3x - 12 = 0 \) gives \( x = 4 \)}

2) \mathrm{\( 3x + 15 = 0 \) gives \( x = -5 \)}

These are the x-intercepts of the two lines. The area between the lines and up to the x-axis is a rectangle with length equal to the distance between the lines and width equal to the difference in x-intercepts.

Using the formula for the distance between parallel lines:\mathrm{\[ d = \frac{{|c1 - c2|}}{{\sqrt{a^2 + b^2}}} = \frac{{|-12 - 15|}}{{\sqrt{3^2 + 4^2}}} = \frac{27}{5} \]}

The width of the rectangle is \mathrm{\( 4 - (-5) = 9 \).}

The area is then \mathrm{\( d \times \text{width} = \frac{27}{5} \times 9 = 48.6 \)} square units.

Tips and Tricks:

1. Identify Parallel Lines: Before calculating areas, always ensure the lines are parallel. Use the coefficients of ( x ) and ( y) to do this.

2. Sketch: A quick sketch of the lines on the graph can give a visual understanding of the area to be calculated.

3. X and Y intercepts: Often, the x or y intercepts of the lines are useful, especially when calculating areas with respect to axes.

4. Use of Previous Concepts: When given equations of lines, always try to deduce if they are parallel, intersecting, or coinciding (from Concept 1). This knowledge will dictate your approach to finding the area.

The understanding of areas between lines and their applications will become clearer with practice. It's crucial to solve various types of questions related to this concept to be well-prepared for the exam.

 

 

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