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Zorluk: ZorLoci and Geometric Constructions

A point P(x,y)P(x, y) moves such that its perpendicular distance from the straight line L1:4x−3y+5=0L_1: 4x - 3y + 5 = 0 is equal to its perpendicular distance from the straight line L2:3x+4y−10=0L_2: 3x + 4y - 10 = 0. Which of the following equations represents one of the straight lines constituting the locus of PP?

  1. x−7y+15=0x - 7y + 15 = 0Cevap
  2. B
    7x−y−5=07x - y - 5 = 0
  3. C
    x+7y+15=0x + 7y + 15 = 0
  4. D
    7x+y+5=07x + y + 5 = 0

Cevap

The equation x−7y+15=0x - 7y + 15 = 0 represents one of the lines constituting the locus.
The locus of a point equidistant from two intersecting straight lines is the pair of angle bisectors between those lines. Setting the perpendicular distance formulas equal yields 4x−3y+5=±(3x+4y−10)4x - 3y + 5 = \pm(3x + 4y - 10). Solving the positive branch yields x−7y+15=0x - 7y + 15 = 0, which correctly represents one of the component lines of the locus.

Adım Adım Çözüm

1
Write the perpendicular distance formulas from point P(x,y)P(x, y) to both given lines.
d1=∣4x−3y+5∣42+(−3)2=∣4x−3y+5∣5d_1 = \frac{|4x - 3y + 5|}{\sqrt{4^2 + (-3)^2}} = \frac{|4x - 3y + 5|}{5} and d2=∣3x+4y−10∣32+42=∣3x+4y−10∣5d_2 = \frac{|3x + 4y - 10|}{\sqrt{3^2 + 4^2}} = \frac{|3x + 4y - 10|}{5}.
The locus of points equidistant from two intersecting lines consists of the angle bisectors of the angles between the lines.
2
Set the two perpendicular distances equal to each other.
\frac{|4x - 3y + 5|}{5} = \frac{|3x + 4y - 10|}{5} \implies |4x - 3y + 5| = |3x + 4y - 10|.
Since the point is equidistant from both lines, d1=d2d_1 = d_2.
3
Remove absolute values by considering both positive and negative cases.
4x−3y+5=±(3x+4y−10)4x - 3y + 5 = \pm(3x + 4y - 10).
Absolute value equality ∣A∣=∣B∣|A| = |B| implies A=BA = B or A=−BA = -B.
4
Evaluate Case 1 (positive sign) to find the first line equation.
4x−3y+5=3x+4y−10  ⟹  (4x−3x)+(−3y−4y)+(5+10)=0  ⟹  x−7y+15=04x - 3y + 5 = 3x + 4y - 10 \implies (4x - 3x) + (-3y - 4y) + (5 + 10) = 0 \implies x - 7y + 15 = 0.
Grouping like terms yields the linear equation for the first angle bisector.
5
Evaluate Case 2 (negative sign) to find the second line equation.
4x−3y+5=−(3x+4y−10)  ⟹  4x−3y+5=−3x−4y+10  ⟹  7x+y−5=04x - 3y + 5 = -(3x + 4y - 10) \implies 4x - 3y + 5 = -3x - 4y + 10 \implies 7x + y - 5 = 0.
Grouping like terms yields the linear equation for the second angle bisector.

Anahtar Kavram

Locus equidistant from two intersecting lines (Angle Bisectors)
Tahmini Süre:2m 0s
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Konular

Loci and Geometric Constructions

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