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Coordinate Geometry - Distance of a Point from a Line

Grade 11ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The perpendicular distance dd is the shortest path from a given point (x1,y1)(x_1, y_1) to a line Ax+By+C=0Ax + By + C = 0. This distance is measured along a line passing through the point that is perpendicular to the given line.

Diagram showing the perpendicular distance d from point P to a straight line.
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The distance of the origin (0,0)(0, 0) from the line Ax+By+C=0Ax + By + C = 0 is simplified to the absolute value of the constant term divided by the square root of the sum of the squares of the coefficients of xx and yy.

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Parallel lines have the same slope. The distance between two parallel lines Ax+By+C1=0Ax + By + C_1 = 0 and Ax+By+C2=0Ax + By + C_2 = 0 is the constant difference ∣C1−C2∣|C_1 - C_2| normalized by the magnitude of the normal vector A2+B2\sqrt{A^2 + B^2}.

Diagram showing two parallel lines L1 and L2 with a perpendicular distance d between them.
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If a point lies on the line, the perpendicular distance is zero, as the coordinates of the point satisfy the equation Ax+By+C=0Ax + By + C = 0.

📐Formulae

Distance dd of point (x1,y1)(x_1, y_1) from line Ax+By+C=0Ax + By + C = 0: d=∣Ax1+By1+C∣A2+B2d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}

Distance dd of the origin (0,0)(0, 0) from line Ax+By+C=0Ax + By + C = 0: d=∣C∣A2+B2d = \frac{|C|}{\sqrt{A^2 + B^2}}

Distance dd between two parallel lines Ax+By+C1=0Ax + By + C_1 = 0 and Ax+By+C2=0Ax + By + C_2 = 0: d=∣C1−C2∣A2+B2d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

Slope mm of the line Ax+By+C=0Ax + By + C = 0: m=−ABm = -\frac{A}{B}

💡Examples

Problem 1:

Find the perpendicular distance of the point P(3,−5)P(3, -5) from the line 3x−4y−26=03x - 4y - 26 = 0.

Solution:

Step 1: Identify the values from the point and the line equation. Here, x1=3x_1 = 3, y1=−5y_1 = -5, A=3A = 3, B=−4B = -4, and C=−26C = -26.

Step 2: Apply the distance formula: d=∣Ax1+By1+C∣A2+B2d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}

Step 3: Substitute the values: d=∣3(3)+(−4)(−5)+(−26)∣32+(−4)2d = \frac{|3(3) + (-4)(-5) + (-26)|}{\sqrt{3^2 + (-4)^2}} d=∣9+20−26∣9+16d = \frac{|9 + 20 - 26|}{\sqrt{9 + 16}} d=∣3∣25d = \frac{|3|}{\sqrt{25}} d=35=0.6d = \frac{3}{5} = 0.6 units.

Explanation:

We use the standard distance formula by plugging in the coordinates of the point into the line's equation in the numerator and dividing by the magnitude of the line's normal vector in the denominator.

Problem 2:

Find the distance between the parallel lines 5x+12y−20=05x + 12y - 20 = 0 and 5x+12y+19=05x + 12y + 19 = 0.

Solution:

Step 1: Identify the coefficients. Since the lines are parallel, A=5A = 5 and B=12B = 12 for both. The constants are C1=−20C_1 = -20 and C2=19C_2 = 19.

Step 2: Apply the formula for the distance between parallel lines: d=∣C1−C2∣A2+B2d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}

Step 3: Substitute the values: d=∣−20−19∣52+122d = \frac{|-20 - 19|}{\sqrt{5^2 + 12^2}} d=∣−39∣25+144d = \frac{|-39|}{\sqrt{25 + 144}} d=39169d = \frac{39}{\sqrt{169}} d=3913=3d = \frac{39}{13} = 3 units.

Explanation:

To find the distance between parallel lines, we calculate the absolute difference between their constant terms and divide by the square root of the sum of the squares of the xx and yy coefficients.

Problem 3:

Find the distance of the origin from the line 4x−3y+10=04x - 3y + 10 = 0.

Graph showing the line 4x - 3y + 10 = 0 and its perpendicular distance from the origin.

Solution:

Given the line 4x−3y+10=04x - 3y + 10 = 0, we have A=4A = 4, B=−3B = -3, and C=10C = 10. The origin is (0,0)(0, 0). Using the formula for distance from origin: d=∣C∣A2+B2d = \frac{|C|}{\sqrt{A^2 + B^2}} d=∣10∣42+(−3)2d = \frac{|10|}{\sqrt{4^2 + (-3)^2}} d=1016+9d = \frac{10}{\sqrt{16 + 9}} d=1025d = \frac{10}{\sqrt{25}} d=105=2d = \frac{10}{5} = 2 Thus, the distance is 22 units.

Explanation:

To find the distance from the origin, we substitute x=0x=0 and y=0y=0 into the numerator of the distance formula, leaving only the constant CC. We then divide by the square root of the sum of the squares of the coefficients of xx and yy.

Problem 4:

Find the distance between the parallel lines y=2x+5y = 2x + 5 and y=2x−3y = 2x - 3.

Graph of two parallel lines y=2x+5 and y=2x-3 illustrating the distance between them.

Solution:

First, rewrite the equations in the form Ax+By+C=0Ax + By + C = 0: Line 1: 2x−y+5=02x - y + 5 = 0 (where A=2,B=−1,C1=5A=2, B=-1, C_1=5) Line 2: 2x−y−3=02x - y - 3 = 0 (where A=2,B=−1,C2=−3A=2, B=-1, C_2=-3) Using the formula for distance between parallel lines: d=∣C1−C2∣A2+B2d = \frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}} d=∣5−(−3)∣22+(−1)2d = \frac{|5 - (-3)|}{\sqrt{2^2 + (-1)^2}} d=∣8∣4+1d = \frac{|8|}{\sqrt{4 + 1}} d=85d = \frac{8}{\sqrt{5}} Rationalizing the denominator: d=855d = \frac{8\sqrt{5}}{5} Thus, the distance is 855\frac{8\sqrt{5}}{5} units.

Explanation:

For parallel lines, we ensure the xx and yy coefficients are identical in both equations. Then, the distance is found by taking the absolute difference of the constants and dividing by the magnitude of the coefficient vector.