Parallel & Perpendicular Lines

Intro

As we've seen in previous lessons, the slope of a line -- the ratio of rise to run -- is measured along perpendicular axes -- also known as orthogonal axes. So, we end up with right triangles in our diagrams -- in which the ratio of the vertical leg length to the horizontal leg length is the slope of the line.

If you've already covered Trigonometry, you'll recognize this ratio as the Tangent of the angle formed between the line and the x-axis.

Here however, we're concerned with similar right triangles formed between the points on the line. Recall that the ratio of the sides of similar figures is constant. So, if the ratio of vertical leg to horizontal leg in one such triangle is 3 : 1, the ratio of all corresponding sides will be 3 : 1. That ratio is the slope of the line. Though we didn't call it slope at the time, we already used this ratio to find the coordinates of a point of division on a given line segment.

This lesson covers two types of lines with related slopes.

intro parallel lines perpendicular lines
examples practice solutions

Parallel Lines

In the diagram, AB // CD.

So triangle EOF ~ triangle AMN; (A, A, A)

therefore

But is the slope of CD,

therefore slope of CD = slope of AB.

Parallel lines have the same slope.

Example

Write the equation of the line:

(a) Through P( – 2, 4), parallel to the line with equation 5x – 3y + 7 = 0.

(b) Through Q( 3, – 1), parallel to the line with equation y = – 7x + 3.

(c) Through R(5, – 3), parallel to the line joining A(6, 0) and B(– 4, 4).

(d) Through the midpoint of LM where L is (–2, 3) and M is (– 8, 7), parallel to y = 4x + 1.

Solution:

(a) 5x – 3y + 7 = 0 has slope of – 5/– 3 or 5/3.

5x - 3y + 22 = 0.

(b) y = – 7x + 3 has a slope of – 7.

y = – 7x + 20 or 7x + y 20 = 0.

(c) The slope of AB is .

2x + 5y 12 = 0.

(d) The midpoint of LM is (–5, 5); the slope of y = 4x + 1 is 4

y – 5 = 4x + 20, which becomes 4x y + 25 = 0.

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PERPENDICULAR LINES:

In the diagram, L1 is perpendicular to L2 and CD is perpendicular to AB. Let DC = q,

AD = p and DB = r. Let m1 = slope L1 and

m2 = slope of L2. So m1 = and m2 =

Now, triangle ADC ~ triangle CDB, so, ,

therefore m1m2 =

therefore

The slopes of perpendicular lines are negative reciprocals.

intro parallel lines perpendicular lines
examples practice solutions

Examples

1) Write the equation of the line through P(–1, 7) perpendicular to the line 7x + 3y – 2 = 0.

Solution:

The slope of 7x + 3y – 2 = 0 is – 7/3, so the slope of a line perpendicular is 3/7.

2) Write the equation of the right bisector of the line segment joining A(3, – 1) and B(5, 3).

(hint: a right bisector is perpendicular to the line, through its midpoint)

Solution:

The midpoint of A(3, – 1) and B(5, 3) is (4, 1). The slope of AB is 2.

3) Prove that the triangle with vertices at A(7, 4) B(2, 3) and C(3, – 2) is a right triangle.

Solution:

3) The slope of AB is 1/5. The slope of BC is 5/–1. Since these are negative reciprocals, the triangle has a right angle at B and so is a right triangle.

4) Prove that the line joining the midpoints of the sides of triangle with vertex A(1, 1) is parallel and equal to one half the base from B(–5, – 1) to C(3, – 9).

Solution:

4)

Length of DE = .

Length of BC = .

therefore DE = ½ BC.

intro parallel lines perpendicular lines
examples practice solutions

Practice

1) Write the equations (in Ax + By + C = 0 form wherever possible) of the line:

2) Prove that A(–10, – 6), B(–7, 1) and C(7, – 5) form a right angled triangle:

3) The midpoints of the adjacent sides of the quadrilateral A(0, 5), B(–3, – 6), C(9, – 10) and D(4, 8) are joined. Test to see if the midpoints are the vertices of a parallelogram.

4) Given the points A(–3, 0), B(–2, – 5) and C(2, 1),

intro parallel lines perpendicular lines
examples practice solutions

Solutions

1) Write the equations (in Ax + By + C = 0 form wherever possible) of the line:

2) Prove that A(–10, – 6), B(-7, 1) and C(7, – 5) form a right angled triangle:

(a) We will find the lengths of AB, BC and AC and show that the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.

(b) Now, we will find the slopes of AB, BC and AC and show that the slope of one is the negative reciprocal of another which will prove that two of the sides are perpendicular to each other.

3) The midpoints of adjacent sides of the quadrilateral A(0, 5), B(–3, – 6), C(9, – 10) and D(4, 8) are joined. Test to see if the midpoints are the vertices of a parallelogram.

E is (, F is (3, – 8), G is , H is

slope EF = , slope GH = , therefore EF // GH

EF =

GH =

EF = GH and EFGH is a parallelogram since a pair of opposite sides are equal and parallel.

4) Given the points A(–3, 0), B(–2, – 5) and C(2, 1). Find point P such that PA // BC, and PC perpendicular to BC.

Let P be (x, y).

Since PA // BC, slope PA = slope BC, so 3x – 2y = – 9

Since PC is perpendicular to BC, slope PC = , so 2x + 3y = 7

We will now solve the system of equations:

3x – 2y = – 9 becomes 6x – 4y = – 18 when we multiply it by 2.

2x + 3y = 7 becomes 6x + 9y = 21 when we multiply it by 3.

and subtracting gives us – 13y = – 39 so y = 3 and x = – 1

The point P is at (–1, 3).

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