Content code
m1315
Slug (identifier)
the-distance-from-a-point-to-a-line-on-a-cartesian-plane
Grades
Secondary IV
Topic
Mathematics
Tags
distance entre point et droite
plan cartésien
périmètre
perpendiculaire
Pythagore
géométrie analytique
pente
taux de variation
Pythagore dans le plan cartésien
distance entre 2 droites
Content
Contenu
Content
Corps

The distance from a point to a line is the length of the shortest possible segment connecting a point to a line.

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2 columns
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First column
Corps

To determine the distance of a point from a line, we measure the length of the perpendicular segment that joins the line to the point.

Second column
Image
A graph showing the distance between the point and the line
Surtitle
Règle
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Corps

Given the equation of the line and the coordinates of the point, the distance between the two can be found by following these steps.

  1. Draw a line passing through point |P| perpendicular to the given line.

  2. Determine the slope of this line: it is the opposite inverse (negative reciprocal) of the slope of the line |d|.

  3. Replace |x| and |y| with the coordinates of |P| to calculate the |y|-intercept of the perpendicular line.

  4. Replace the found values of |m| and |b| to get the equation of the perpendicular line.

  5. Use the comparison method to find the point of intersection between |d| and the perpendicular line (if necessary, see solving a system of equations).

  6. Apply the formula for the distance between two points to find the distance between the point |P| and the line |d|.

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Corps

What is the distance between the line |d| with the equation |y = -2x+7| and the point |P| with coordinates |(6,4)|?

Solution
Corps

The animation below shows the steps in detail.

Corps

These steps can be simplified by using a formula to calculate the distance between a point and a line.

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Corps

Slope-intercept equation:
Given a point |P(x_1, y_1)| and a line |D| with the equation |y = mx + b|, use the following formula.

|d(D,P) = \displaystyle \frac{\mid mx_1 - y_1 + b \mid}{\sqrt{m^2 + 1}}|

Standard form equation:
Given a point |P(x_1,y_1)| and a line |D| with the equation |Ax + By + C = 0|, use the following formula.

|d(D,P) = \displaystyle \frac{\mid Ax_1 + By_1 + C \mid}{\sqrt{A^2 + B^2}}|

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2 columns
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Corps

What is the distance between the line with equation |y = 3x – 4| and the point P with coordinates |(4 , -1)|?

Second column
Image
A graph showing the distance between the point and the line
Solution
Corps

Since the line is written in the slope-intercept form, we will use the first formula. ||\begin{align} d &= \dfrac{\mid mx_1 - y_1 + b\mid}{\sqrt{m^2 + 1}} \\ d &= \dfrac{\mid (3\times 4) - (-1) + (-4)\mid}{\sqrt{3^2 + 1}} \\d &= \dfrac{\mid 12 + 1 - 4\mid}{\sqrt{9 + 1}} = \frac{\mid 9 \mid}{\sqrt{10}} \approx 2.85\ \text{units} \end{align}||

The exact same result can be obtained by using the other formula.
Transform the equation of the line into standard form: |0 = 3x – y - 4.| ||\begin{align} d &= \dfrac{\mid Ax_1 + By_1 + C \mid}{\sqrt{A^2 + B^2}} \\ d &= \dfrac{\mid (3\times 4) + (-1 \times -1) + (-4) \mid}{\sqrt{3^2 + (-1)^2}} \\ d &= \dfrac{\mid 12 + 1 - 4 \mid}{\sqrt{9 + 1}} \end{align}||

This gives the same result, i.e. approximately |2.85| units.

Title (level 2)
Exercise
Title slug (identifier)
exercices
Contenu
Contenu
Title
The Distance From a Point to a Line
Title slug (identifier)
the-distance-from-a-point-to-a-line
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