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Chapter 10: Problem 61

Suppose that the points \((h, k),(1,2)\) and \((-3,4)\) lie on the line \(\mathrm{L}_{1}\). If a line \(\mathrm{L}_{2}\) passing through the points \((h, k)\) and \((4,3)\) is perpendicular on \(\mathrm{L}_{1}\), then equals : [April08, 2019 (II)] (a) \(\frac{1}{3}\) (b) 0 (c) 3 (d) \(-\frac{1}{7}\)

Short Answer

Expert verified
The value of \( h \) is 3.

Step by step solution

01

Find the Slope of Line L1

To find the slope of line \( L_1 \), we use the formula for the slope between two points: \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Using the points \((1, 2)\) and \((-3, 4)\), the slope \( m_1 \) is calculated as: \[ m_1 = \frac{4 - 2}{-3 - 1} = \frac{2}{-4} = -\frac{1}{2} \].
02

Determine the Slope of Line L2

Lines that are perpendicular have slopes that are negative reciprocals. Therefore, if \( m_1 = -\frac{1}{2} \), the slope of line \( L_2 \), denoted \( m_2 \), must satisfy: \[ m_2 \times m_1 = -1 \]. Thus, \[ m_2 = 2 \].
03

Use the Slope of Line L2 to Find (h, k)

Given that line \( L_2 \) passes through \((h, k)\) and \((4, 3)\) with a slope \( m_2 = 2 \), we use the slope formula: \[ m_2 = \frac{3 - k}{4 - h} \]. Setting \( 2 = \frac{3 - k}{4 - h} \), we can solve for \( k \) and \( h \).
04

Solve for k in terms of h

Rearrange the equation \( 2 = \frac{3 - k}{4 - h} \) as: \[ 2(4 - h) = 3 - k \]. This gives \[ 8 - 2h = 3 - k \] or \[ k = 3 - (8 - 2h) \]. Simplifying, we find \[ k = 2h - 5 \].
05

Ensure (h, k) Lies on Line L1

Since \((h, k)\) lies on line \( L_1 \), it must satisfy the equation derived from its slope; using point \((1, 2)\), the equation is \( y - 2 = -\frac{1}{2}(x - 1) \). Substituting \( h \) and \( k = 2h - 5 \), we get: \[ (2h - 5) - 2 = -\frac{1}{2}(h - 1) \].
06

Simplify and Solve for h

Simplify the equation: \[ 2h - 7 = -\frac{1}{2}(h - 1) \]. Distribute: \[ 2h - 7 = -\frac{h}{2} + \frac{1}{2} \]. Multiply through by 2 to clear the fraction: \[ 4h - 14 = -h + 1 \]. Rearrange to solve for \( h \): \[ 5h = 15 \], thus: \[ h = 3 \].
07

Find the Value of k

Substitute \( h = 3 \) back into the equation \( k = 2h - 5 \): \[ k = 2(3) - 5 = 6 - 5 = 1 \].
08

Conclusion

The value of \((h, k)\) that satisfies all conditions is \((3, 1)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equation of a Line
In analytical geometry, a line can be uniquely described by its equation, which relates the coordinates of every point on the line. One common form of the equation of a line is the slope-intercept form, given by the equation: \[ y = mx + b \]
  • Here, \( m \) represents the slope of the line, which measures its steepness or angle relative to the x-axis.
  • The term \( b \) is the y-intercept, which is the point where the line crosses the y-axis.
Another commonly used form is the point-slope form:\[ y - y_1 = m(x - x_1) \]This form is particularly useful when you know a point on the line, \((x_1, y_1)\), and the slope \( m \). Understanding these forms allows you to write the equation of a line when given sufficient information like the slope and a point.
Slope of a Line
The slope is a crucial concept in understanding how lines behave in geometry. It is defined as the change in the y-coordinate divided by the change in the x-coordinate between two distinct points on a line. This is often represented as:\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]The slope tells us several things about a line:
  • A positive slope indicates the line rises from left to right, whereas a negative slope indicates it falls.
  • A slope of zero means the line is horizontal, and an undefined slope (division by zero) means the line is vertical.
This simple ratio encapsulates the direction and steepness of the line. Hence, understanding slope is foundational to the ability to describe and analyze linear relationships in geometry.
Perpendicular Lines
Perpendicular lines form a specific angle of 90 degrees between each other. In terms of geometry and algebra, two lines are said to be perpendicular if the product of their slopes is \(-1\). Thus, if a line \( L_1 \) has a slope \( m_1 \), then a line \( L_2 \) is perpendicular to it if its slope \( m_2 \) satisfies the condition:\[ m_1 \times m_2 = -1 \]
  • For example, if a line has a slope of \( -\frac{1}{2} \), then a perpendicular line will have a slope of \( 2 \) because \(-\frac{1}{2} \times 2 = -1\).
This relationship helps design and verify geometric structures where right angles are necessary and is pivotal in multiple applications, such as creating grid plans and navigational designs.

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Most popular questions from this chapter

The foot of the perpendicular drawn from the origin, on the line, \(3 x+y=\lambda(\lambda \neq 0)\) is \(P\). If the linemeets \(x\)-axisat \(A\) and \(y\)-axis at \(B\), then the ratio \(B P: P A\) is [Online April \(15, \mathbf{2 0 1 8}]\) (a) \(9: 1\) (b) \(1: 3\) (c) \(1: 9\) (d) \(3: 1\)

Locus of mid point of the portion between the axes of \(x \cos \alpha+y \sin \alpha=p\) whre \(p\) is constant is [2002] (a) \(x^{2}+y^{2}=\frac{4}{p^{2}}\) (b) \(x^{2}+y^{2}=4 p^{2}\) (c) \(\frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{2}{p^{2}}\) (d) \(\frac{1}{x^{2}}+\frac{1}{y^{2}}=\frac{4}{p^{2}}\)

A straight line through origin \(\mathrm{O}\) meets the lines \(3 \mathrm{y}=10-4 \mathrm{x}\) and \(8 x+6 y+5=0\) at points A and B respectively. Then \(O\) divides the segment \(\mathrm{AB}\) in the ratio: [Online April 10, 2016] (a) \(2: 3\) (b) \(1: 2\) (c) \(4: 1\) (d) \(3: 4\)

Two vertical poles of heights, \(20 \mathrm{~m}\) and \(80 \mathrm{~m}\) stand apart on a horizontal plane. The height (in meters) of the point of intersection of the lines joining the top of each pole to the foot of the other, from this horizontal plane is : [April08, 2019 (II)] (a) 15 (b) 18 (c) 12 (d) 16

If the line \(3 x+4 y-24=0\) intersects the \(x\)-axis at the point A and the \(y\)-axis at the point \(B\), then the incentre of the triangle OAB, where \(\mathrm{O}\) is the origin, is: [Jan. \(\mathbf{1 0}, \mathbf{2 0 1 9}\) (I)] (a) \((3,4)\) (b) \((2,2)\) (c) \((4,3)\) (d) \((4,4)\)
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