FIND THE DISTANCE FROM A POINT TO THE LINE

The distance from the point to the line is the distance from the point to the foot of the perpendicular.

Find the distance from :

Example 1 :

(7, - 4) to y  =  3x - 5

Solution :

Step 1 :

y  =  3x–5  -----(1)

By comparing the given equation with slope intercept form (y = mx+b).

slope (m)  =  3

Step 2 :

Slope of perpendicular line  =  -1/3

Step 3 :

Find the equation of the perpendicular line passes through (7, -4).

(7, -4) ----->(x₁, y₁)

m  =  -1/3

(y–y₁)  =  m(x–x₁)

(y+4)  =  - 1/3(x–7)

3(y+4)  =  -1(x-7) 

3y  =  -x+7-12

3y  =  -x-5  ---(2)

Step 4 :

Find the point of intersection of two lines,

By applying (1) in (2), we get

Foot of perpendicular is (1, -2).

Step 5 :

Distance between two points :

=  √(1–7)² + (-2+4)²

=  √40

=  2√10 units.

So, the required distance is 2√10 units.

Example 2 :

(-6, 0) to y = 3–2x

Solution :

y  =  3–2x  ----(1)

By comparing the given equation with y = mx+b, we come so know slope (m)

Slope (m)  =  - 2

Slope of perpendicular line  =  1/2

Equation of perpendicular line :

(y - 0)  =  1/2(x + 6)

2y  =  x+6  ----(2)

Applying (1) in (2), we get

So, foot of perpendicular is (0, 3).

Distance between (-6, 0) and (0, 3).

=  √(0+6)² + (3-0)²

=  √(36+9)

=  √45

=  3√5 units.

So, the required distance is 3√5 units.

Example 3 :

Shown on the grid is the line, y = x +1

(a) Find the gradient of a line perpendicular to y = x + 1

(b) Find the equation of the line perpendicular to y = x + 1 that passes through the point (5, 0)

(c) Find the point where your answer to (b) and the line y = x + 1 intersect.

(d) Calculate the distance between your answer to (c) and the point (5, 0)

Solution :

(a) Slope of the line y = x + 1

Comparing with y = mx + b, we know that m = 1.

(b) Slope of the line y = x + 1 is 1. since the required line is perpendicular to the required line, we have to use -1/m to find the slope of the perpendicular line.

-1 is the slope and passes through the point (5, 0).

y - 0 = -1(x - 5)

y = -x + 5

c) Finding point of intersection of the line y = x + 1 and y = -x  + 5.

x + 1 = -x + 5

x + x = 5 - 1

2x = 4

x = 2

Applying x = 2 in y = x + 1, we get y = 3

So, the required point of intersection is (2, 3).

d) Distance between the point (2, 3) and (5, 0).

=  √(5 - 2)² + (3 - 0)²

=  √3² + 3²

= √(9 + 9)

= √18

= √(2 x 3 x 3)

= 3√2 units.

So, the required distance is 3√2 units.

Example 4 :

• The line L1 has equation y = 2x - 4
• The line L2 has equation y = -3x + 11
• The line L3 has equation y = x
• L1 and L2 meet at the point A.
• Line 1 and line 3 are intersecting at B.

Work out the shortest distance between the points A and B.

Solution :

Let us find the point of intersection of L1 and L2.

y = 2x - 4  ------(1)

y = -3x + 11  -----(2)

(1) = (2)

2x - 4 = -3x + 11

2x + 3x = 11 + 4

5x = 15

x = 15/5

x = 3

Applying x = 3 in (1), we get

y = 2(3) - 4

y = 6 - 4

y = 2

So, the point of intersection of line 1 and line is A (3, 2).

y = 2x - 4 -----(3)

y = x -----(4)

x = 2x - 4

x - 2x = -4

-x = -4

x = 4

Applying x = 4 in (4)

y = 4

So, point B is (4, 4).

AB = √(4 - 3)² + (4 - 2)²

= √1² + 2²

= √(1+4)

= √5 units.

Calculate the shortest distance between the point G(-4, 4) and the line y = 3x- 4.

Solution :

Step 1 :

Determine the equation of the line passing through (-4, 4) and perpendicular to y = 3x- 4.

Slope of the given line y = 3x - 4 is 3

Slope of the perpendicular line :

-1/m = -1/3

Equation of the perpendicular line is 

y - y₁ = -1/m(x - x₁)

y - 4 = -1/3(x + 4)

y = (-1/3) x - (4/3) + 4

y = (-1/3) x + (8/3)

Step 2 :

Solving y = 3x - 4 and y = (-1/3) x + (8/3)

3x - 4 = (-1/3) x + (8/3)

3x + x/3 = 8/3 + 4

10x / 3 = 20/3

10x = 20

x = 2

Applying x = 2, we get

y = 3(2) - 4

y = 2

So, (2, 2) is the required point. The distance between (-4, 4) and (2, 2).

Step 3 :

= √(-4 - 2)² + (4 - 2)²

= √(-6)² + (2)²

= √(36 + 4)

= √40

= 2√10 units

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