/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 79 A straight line through the poin... [FREE SOLUTION] | 91Ó°ÊÓ

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A straight line through the point \(A(3,4)\) is such that its intercept between the axes is bisected at \(A\). Its equation is (A) \(x+y=7\) (B) \(3 x-4 y+7=0\) (C) \(4 x+3 y=24\) (D) \(3 x+4 y=25\)

Short Answer

Expert verified
Option (C) \(4x + 3y = 24\).

Step by step solution

01

Understanding the Problem

We need to find the equation of a line that passes through the point \( A(3,4) \) and is such that its intercepts on the x and y axes are bisected at \( A \).
02

Representing Intercepts with Variables

Let the x-intercept be \( p \) and the y-intercept be \( q \). Then, the equation of the line can be written in intercept form as \( \frac{x}{p} + \frac{y}{q} = 1 \).
03

Using the Bisect condition

Since the intercepts are bisected at \( A(3,4) \), the midpoint of the intercepts is \((\frac{p}{2}, \frac{q}{2}) = (3,4) \). Hence, \( \frac{p}{2} = 3 \) and \( \frac{q}{2} = 4 \).
04

Solving for Intercepts

From \( \frac{p}{2} = 3 \), we get \( p = 6 \). From \( \frac{q}{2} = 4 \), we get \( q = 8 \). So, the intercepts are 6 on the x-axis and 8 on the y-axis.
05

Writing the Line Equation

Using the intercepts \( p = 6 \) and \( q = 8 \), the equation becomes \( \frac{x}{6} + \frac{y}{8} = 1 \). Multiplying through by 24 to eliminate fractions gives \( 4x + 3y = 24 \).
06

Choosing the Correct Option

The equation \( 4x + 3y = 24 \) corresponds to option (C).

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Key Concepts

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

Intercept Form
The intercept form of a line equation is a convenient way to express lines based on their intersections with the coordinate axes. This form is particularly useful when you know where a line crosses the x-axis and y-axis, known as the x-intercept and y-intercept, respectively.
The intercept form is expressed as:
  • \( \frac{x}{p} + \frac{y}{q} = 1 \)
where:\( p \) is the x-intercept,
\( q \) is the y-intercept.
Essentially, \( p \) and \( q \) represent the distances from the origin to the point where the line intersects the respective axis. If you know the intercepts, this form offers a straightforward way to derive the line's equation.
This format is practical for various geometrical scenarios where intercepts are primary data, leading to quick solutions.
Bisecting Line Intercepts
When we say a line bisects the intercepts, it means that a specific point, in this case, point \( A(3,4) \), is the midpoint of those intercepts. Here, the term 'intercepts' refers to the parts of the line that lie between the points where it crosses the x-axis and y-axis.
To find these intercepts, think about the midpoint formula, which combines the x and y coordinates and divides by two to find the middle point. In this problem, this method finds the point \( A \) in relation to the intercepts:
  • The midpoint is given as \( (\frac{p}{2}, \frac{q}{2}) = (3,4) \), representing coordinates halfway between the intercepts.

  • Given \( \frac{p}{2} = 3 \), solve for \( p \) to find the x-intercept, so \( p = 6 \).

  • Likewise, \( \frac{q}{2} = 4 \) gives the y-intercept, so \( q = 8 \).
This approach exploits the relationship between midpoints and endpoints to find precise line segments.
Point-Slope Form
The point-slope form is another essential technique in coordinate geometry for defining a line. It is particularly useful when you are given a point on the line and the line's slope, or gradient. The point-slope form is written as:
  • \( y - y_1 = m(x - x_1) \)
where:
\( (x_1, y_1) \) is a specific point on the line,
\( m \) is the slope of the line.

In this exercise, we translated the problem into intercepts and used different form representations, but knowing point-slope form helps when those intercepts or coordinates need further elucidation, like finding parallel lines, tangent slopes, or perpendicular bisectors based on similar data.
Coordinate Geometry
Coordinate geometry, or analytic geometry, allows us to ascertain and solve geometric problems using a coordinate plane. We use algebraic formulas to describe curves and shapes with numeric equations.
In coordinate geometry, any line can be defined with key attributes like slope, intercepts, and specific points it passes through. This realm of math is powerful because it bridges algebra and geometry, turning geometrical problems into algebraic ones.
  • Pass through points to define shapes.
  • Derive equations to express geometrical principles.
  • Apply midpoints, distances, and slopes as essential tools.
Such geometry is integral to solving complex situational problems like optimizing shapes and movements within defined space, enhancing clarity and understanding of spatial relationships.

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Most popular questions from this chapter

\(O X\) and \(O Y\) are two coordinate axes. On \(O Y\) is taken a fixed point \(P\) and on \(O X\) any point \(Q .\) On \(P Q\) an equilateral triangle is described, its vertex \(R\) being on the side of \(P Q\) away from \(O\), then the locus of \(R\) will be (A) straight line (B) circle (C) ellipse (D) parabola

A line which makes an acute angle \(\theta\) with the positive direction of \(x\)-axis is drawn through the point \(P(3,4)\) to meet the line \(x=6\) at \(R\) and \(y=8\) at \(S\), then (A) \(P R=3 \sec \theta\) (B) \(P S=4 \operatorname{cosec} \theta\) (C) \(P R+P S=\frac{2(3 \sin \theta+4 \cos \theta)}{\sin 2 \theta}\) (D) \(\frac{9}{(P R)^{2}}+\frac{16}{(P S)^{2}}=1\)

\(A(0,0), B(2,1)\) and \(C(3,0)\) are the vertices of a \(\triangle A B C\) and \(B D\) is its altitude. If the line through \(D\) parallel to the side \(A B\) intersects the side \(B C\) at a point \(K\), then the product of the areas of the triangles \(A B C\) and \(B D K\) is (A) 1 (B) \(\frac{1}{2}\) (C) \(\frac{1}{4}\) (D) none of these

In oblique coordinates, the equation \(y=m x+c\) represents a straight line which is inclined at an angle $$ \tan ^{-1}\left(\frac{m \sin w}{1+m \cos w}\right) $$ to the \(x\)-axis, where \(w\) is the angle between the axes. If \(\theta\) be the angle between two lines \(y=m_{1} x+c_{1}\) and \(y=m_{2} x\) \(+c_{2}, w\) be the angle between the axes, then $$ \tan \theta=\frac{\left(m_{1}-m_{2}\right) \sin w}{1+\left(m_{1}+m_{2}\right) \cos w+m_{1} m_{2}} $$ The two given lines are parallel if \(m_{1}=m_{2}\). The two lines are perpendicular if \(1+\left(m_{1}+m_{2}\right) \cos w+\) \(m_{1} m_{2}=0\) If the straight lines \(y=m_{1} x+c_{1}\) and \(y=m_{2} x+c_{2}\) make equal angles with the axis of \(x\) and be not parallel to one another, then \(m_{1}+m_{2}+k m_{1} m_{2} \cos w=0\) where \(k=\) (A) 1 (B) 2 (C) \(-1\) (D) \(-2\)

If the point \((2 \cos \theta, 2 \sin \theta)\) does not fall in that angle between the lines \(y=|x-2|\) in which the origin lies then \(\theta\) belongs to (A) \(\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)\) (B) \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) (C) \((0, \pi)\) (D) none of these

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