Examples based on Straight Line

Straight Lines is an important topic of coordinate geometry. Various competitive exams like the JEE usually ask several questions from this topic. Students planning to take such elite exams must try to cover all possible topics of IIT JEE Mathematics including straight lines in order to remain competitive in the JEE.   

Illustration 1:   

where, m = slope = 3/4  

Intercept Form  

We can write the given line as 3x – 4y = –5  

⇒ 3x/–5 + 4y/5 = 1  

x/–5/3 + y/5/4 = 1 (intercept form)  

x – intercept = –5/3 

y – intercept = 5/4
 

y – 2 = 3/4 (x – 1)
 

Parametric Form 

Normal Form

3x – 4y = – 5  

–3x + 4y = 5

Dividing by ((–3)² + 4)^½    

⇒ –3/5 x + 4/5 y = 5/5 = 1  

⇒ x cos α + y sin α = p, p > 0  

Where cos α = –3/5, sin α = 4/5, p = 1

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Illustration 2:

Equation of the line through the point P is

Method 2. 

Equation of the line through the point P is y = 1/√3 x + 3 [because here m = tan 30^o = –1/√3, c = 3]

Solving this line, with x + y = 6, we get

xQ = 3√3/√3+1, yQ = 3√3+6/√3+1 

distance PQ = √(xQ – xP)² + (yQ – yP)²

= 6/√3+1

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Illustration 3: 

Find the equation of the line whose perpendicular distance from the origin is 4 units and the angle which the normal makes with positive direction of x-axis is 15^o.

Solution:   

Here, we are given p = 4 and ω = 15^o.

Now cos 15^o = √3+1/2√2

and sin 15^o = √3–1/2√2

By the normal form, the equation of the line is

x cos 15^o + y sin 15^o = 4 or √3+1/2√2 x + √3–1/2√2 y = 4

or (√3 + 1) x + (√3 – 1)y = 8√2.

This is the required equation.

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Illustration 4:

Find the equation to the straight line which passes through the point (–5, 4) and is such that the position of it between the axes is divided by the given point in the ratio 1 : 2. 

Solution:

Let the required straight line be (x/a) + (y/b) = 1.  

Using the given conditions, P (2a+1.0/2+1, 2.0+1.b/2+1) is the point which divides (a, 0) and (0, b) internally in the ratio 1 : 2.

Hence –5 = 2a/3, 4 = b/3 ⇒ a = –15/2, b = 12.

⇒ tan θ = –√8 = slope of line.

We know that the equation of the straight line passing through the point (x₁, y₁) having slope m is y – y₁ = m(x – x₁).

Therefore the equation of the required line is y – 2 = –√8 (x – 1)

⇒ √8 x + y – √8 – 2 = 0.

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Illustration 5:

Find the equation of the line joining the points (–1, 3) and (4, –2).

Solution:

Equation of the line passing through the points (x₁, y₁) and (x₂, y₂) is

y – y₁ = y₁–y₂/x₁–x₂ (x – x₁)

Hence equation of the required line will be

y – 3 = 3+2/–1–4 (x + 1) ∝ x + y – 2 = 0.

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Illustration 6:

It is possible to find the distance of of a point form a line but how can we determine as to which side of the line does the point lie?

Solution:

From the figure, we observe that

ax₀ + by₀ + c = 0     (Since the point (x₀, y₀) lies on the line)        …… (1)

Consider, ax₁ + by₁ + c

        = (ax₀ + c) + by₁                [x₀ = x₁ = x₂]

        = b(y₁ – y₀)

        = –ve

Consider, ax₂ + by₂ + c

        = (ax₀ + c) + by₂

        = b(y₂ – y₀)

        = +ve

Thus we observe that the point is on one side of the line, if put in the expression of line is gives one sign, while the point is on the other side of the line, if put in the expression of line it gives opposite sign.

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Illustration 7: 

Represent the straight-line y = x + 2 in the parametric form.

Solution: 

Slope of the given line is = 1 = tan π/4.

Equation of the straight line can be written as y – 2 = x.

or y–2/1/√2 = x/1/√2= r.

Any point on the line is (r/√2, 2 + r/√2).

The point (x, y) is at a distance r from the point (0, 2).

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Illustration 8:

Tangents are drawn from the point P(3, 4) to the ellipse x²/9 + y²/4 = 1 touching the ellipse at the points A and B. The equation of the locus of the point whose distances from the point P and the line AB are equal is...? (IIT JEE 2010)

Solution:

Clearly, the moving point traces a parabola with focus at P(3, 4) and directrix as AB:

(y-0)/(x-3) = -1/3 or x+3y -3 = 0

Hence, the equation of parabola is

(x-3)² + (y-4)² = (x+3y-3)²/10

Hence, the required equation is

9x² + y² – 6xy – 54x – 62y + 241 = 0.

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Illustration 9:

A straight line moves so that the sum of the reciprocals of its intercepts on the coordinate axis is constant. Show that it passes through a fixed point.

Solution:

Let the variable line be x/a + y/b = 1.

Then, we are given in the question that 1/a + 1/b = K, where K is a constant.

Hence, the variable line takes the form

x/a + (K-1/a)y – 1 = 0

or 1/a(x-y) + (Ky-1) = 0

This clearly represents a straight line through the intersection of x-y = 0 and Ky – 1 = 0.

The lines intersect at (1/K, 1/K) and hence x/a + y/b = 1 always passes through the fixed point (1/K, 1/K).

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Illustration 10:

Find the acute angle between the lines 

2x + y + 11 = 0, x – 6y + 7 = 0.

Solution:

The gradients of the above lines are -2 and 1/6. 

Also, the angle of slope of the first line is in second quadrant while that of second line is in first quadrant. Hence, we have

m₂ = -2, m₁ = 1/6 and tan θ = (-2-1/6)/(1 + (-2)(1/6)) = -13/4

This shows that θ is an obtuse angle. 

If φ is the acute angle between the lines, then θ = 180^o – φ 

and so we have, tan θ = tan (180^o – φ).

But, tan (180^o – φ) = – tan φ and so

 tan θ = – tan φ

So, we have tan φ = - tan θ = -(-13/4) = 13/4.

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Illustration 11: 

A line joining two points A(2, 0) and B(3, 1) is rotated about A in the anticlockwise direction through an angle of 15^o. Find the equation of the line in the new position. If B goes to C, what will be the coordinates of C, in the new position?

Solution:

Slope of BAB(m) = 1 ⇒ m = tan θ = 1 ⇒ θ = 45^o.

= tan (60^o) (because angle between AB and AC = 15^o).

Also AB = AC = √2 and A is (2, 0).

Hence equation of the line AC is

(x–2)/cos 60^o = (y–0)/sin 60^o

⇒ C is (2 + √2.1/2, 0 + √2.√3/2) i.e. C is (2 + 1/√2, √3/√2).

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Illustration 12:

Find the distance of the point P(–2, 3) from the line AB which is x – y = 5.

Solution:

The equation of the line is

x – y – 5 = 0 [Making right side zero]

∴ Perpendicular Distance of the point (2, 3)

= (–2)–3–5/√(1)²+(–1)² = –10/√2

= –5√2

∴ Changing the sign, perpendicular Distance in magnitude = 5√2.

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Illustration 13:

Draw the lines 3x + 4y – 12 = 0 and 5x + 12y + 13 = 0. Find the equation of the bisector of the angle containing the origin. Also find the acute angle bisector and obtuse angle bisector.