Use these Inter 1st Year Maths 1B Formulas PDF Chapter 3 The Straight Line to solve questions creatively.

## Intermediate 1st Year Maths 1B The Straight Line Formulas

→ Equation of a horizontal line is y = k

→ Equation of a vertical line is x = h

→ If ‘θ ‘ is the inclination of a line then slope = m = tan θ

→ If m_{1} m_{2} are slopes of two lines and ‘θ ‘ is angle between them then tan θ = \(\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\)

→ If two lines are parallel then their slopes are equal i.e., m_{1} = m_{2}

→ If two lines are perpendicular, then m_{1}m_{2} = -1

→ Equation of a line passing through (x_{1}, y_{1}) with slope ‘m’ is y – y_{1} = m(x – x_{1})

→ Equation of a line passing through origin with slope m is y = mx

→ Equation of the line passing through (x_{1}, y_{1}) and (x_{2}, y_{2}) is \(\) (or) (x – x_{1})(y_{1} – y_{2}) = (y – y_{1})(x_{1} – x_{2})

→ Equation of the line with slope’m’ and having

- y-intercept ‘c’ is y = mx + c (slope intercept form)
- x-intercept ‘a’ is y = m(x – a)

→ Equation of a line in intercept form is \(\frac{x}{a}+\frac{y}{b}\) = 1

- Intersecting point on X – axis = (a, 0)
- Intersecting point on Y – axis = (0, b)

→ Area of Δ^{le} formed by the line with co-ordinate axis is = \(\frac{1}{2}\)|ab|

→ Area of Δ^{le} formed by the line ax + by + c = 0 with co-ordinate axis is \(\frac{c^{2}}{2|a b|}\)

→ If a point (x_{1}, y_{1}) divides the line segment between the co-ordinate axes in the ratio l: m then equation of the line is \(\frac{m x}{x_{1}}+\frac{l y}{y_{1}}\) = l + m

→ Equation of a line in normal form or perpendicular form is x cos α + y sin α = p where ‘α’ is the angle made by the perpendicular with +ve X – axis. ‘p’ is length of normal form origin to the line.

→ Perpendicular distance from (x_{1}, y_{1}) to the line ax + by + c = 0 is \(\frac{\left|a x_{1}+b y_{1}+c\right|}{\sqrt{a^{2}+b^{2}}}\)

→ Perpendicular distance from origin to the line ax + by + c = 0 is \(\frac{|c|}{\sqrt{a^{2}+b^{2}}}\)

→ The ratio in which the line L ax + by + c = 0 (ab ≠ 0) divides the line segment AB joining the points A(x_{1}, y_{1}) and B(x_{2}, y_{2}) is –\(\left(\frac{a x_{1}+b y_{1}+c}{a x_{2}+b y_{2}+c}\right)\) or –\(\frac{L_{11}}{L_{22}}\)

where L_{11} = a_{1}x + b_{1}y + c; L_{22} = a_{2}x + b_{2}y + c

→ If L_{11}, L_{22} are having same sign or opposite sign’s then those points lies same side or opposite sides of the line L = 0 respectively.

→ If the foot of the perpendicular from (x_{1}, y_{1}) to the line ax + by + c = 0 is (α, β) then \(\frac{\alpha-x_{1}}{a}=\frac{\beta-y_{1}}{b}=\frac{-\left(a x_{1}+b y_{1}+c\right)}{a^{2}+b^{2}}\)

→ If the image of(x_{1}, y_{1}) w.r.t. the line ax +by + c = 0 is (α, β) then

\(\frac{\alpha-x_{1}}{a}=\frac{\beta-y_{1}}{b}=\frac{-2\left(a x_{1}+b y_{1}+c\right)}{a^{2}+b^{2}}\)

→ The point of intersection of two lines a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0 is

\(\left(\frac{b_{1} c_{2}-b_{2} c_{1}}{a_{1} b_{2}-a_{2} b_{1}} \cdot \frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}\right)\)

→ If θ |0 ≤ θ ≤ π| is angle between the lines a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0 then cos θ = \(\frac{a_{1} a_{2}+b_{1} b_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}}}\)

sin θ = \(\frac{a_{1} b_{2}-a_{2} b_{1}}{\sqrt{a_{1}^{2}+b_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}}}\)

and tan θ = \(\frac{a_{1} b_{2}-a_{2} b_{1}}{a_{1} a_{2}+b_{1} b_{2}}\)

- The lines are perpendicular ⇔ a
_{1}a_{2}+ b_{1}b_{2}= 0 - Then lines are parallel ⇔ \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\)

→ The equation of a line passing through the point (x_{1}, y_{1}) and parallel to the line ax + by + c = 0 is a(x – x_{1}) + b(y – y_{1}) = 0

→ The equation of a line passing through the point (x_{1}, y_{1}) and perpendicular to ax + by + c = 0 is b(x – x_{1}) – a(y – y_{1}) = 0

→ The distance between the two parallel straight lines a_{1}x + b_{1}y + c_{1} = 0 and a_{1}x + b_{1}y + c_{2} = 0 is \(\frac{\left|c_{1}-c_{2}\right|}{\sqrt{a_{1}^{2}+b_{1}^{2}}}\)

→ Area of the parallelogram formed by the lines a_{1}x + b_{1}y + c_{1} = 0,a _{2}x + b_{2}y + d_{1}= 0, a_{2}x + b_{2}y + c_{2} = 0 and a_{2}x + b_{2}y + d_{2} = 0 is \(\left|\frac{\left(d_{1}-c_{1}\right)\left(d_{2}-c_{2}\right)}{a_{1} b_{2}-a_{2} b_{1}}\right|\)

→ Equation of the line parallel to X – axis and passing through (x_{1}, y_{1}) is y = y_{1}

→ Equation of the line parallel to Y- axis and passing through (x_{1}, y_{1}) is x = x_{1}

**Concurrent lines-properties related to a Triangle
**Theorem:

The medians of a triangle are concurrent.

Proof:

Let A(x

_{1}, y

_{1}), B(x

_{2}, y

_{2}), C(x

_{3}, y

_{3}) be the vertices of the triangle

Let D,E,F be the mid points of BC, CA, AB respectively

∴ D = \(\left(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}\right)\)

E = \(\left(\frac{x_{3}+x_{1}}{2}, \frac{y_{3}+y_{1}}{2}\right)\)

F = \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\)

Slope of \(\overline{A D}\) is \(\frac{\frac{y_{2}+y_{3}}{2}-y_{1}}{\frac{x_{2}+x_{3}}{2}-x_{1}}=\frac{y_{2}+y_{3}-2 y_{1}}{x_{2}+x_{3}-2 x_{1}}\)

Equation of \(\overline{A D}\) is

y – y_{1} = \(\frac{y_{2}+y_{3}-2 y_{1}}{x_{2}+x_{3}-2 x_{1}}\)(x – x_{1})

⇒ (y – y_{1})(x_{2} + x_{3} – 2x_{1}) = (x – x_{1})(y_{2} + y_{3} – 2y_{1})

⇒ L_{1} = (x – x_{1})(y_{2} + y_{3} – 2y_{1}) – (y – y_{1})(x_{2} + x_{3} – 2x_{1}) = 0

Similarly, the equations to \(\overline{B E}\) and \(\overline{C F}\) respectively are L_{2} s (x – x_{2})(y_{3} + y_{1} – 2y_{2}) – (y – y_{2}) (x_{3} + x_{1} – 2x_{2}) = 0.

L_{3} = (x – x_{3})(y_{1} + y_{2} – 2y_{3}) – (y – y_{3}) (x_{1} + x_{2} – 2x_{3}) = 0.

Now 1. L_{1} + 1.L_{2} + 1. L_{3} = 0

The medians L_{1} = 0, L_{2} = 0, L_{3} = 0 are concurrent.

Theorem:

The altitudes of a triangle are concurrent.

Proof:

Let A(x_{1}, y_{1}), B(x_{2}, y_{2}), C(x_{3}, y_{3}) be the vertices of the triangle ABC. Let AD, BE,CF be the altitudes.

Slope of \(\overline{B C}\) is \(\frac{y_{3}-y_{2}}{x_{3}-x_{2}}\) and AD ⊥ BC

Slope of the altitude through A is – \(\frac{x_{3}-x_{2}}{y_{3}-y_{2}}\)

Equation of the altitude through A is y – y_{1} = \(\frac{x_{3}-x_{2}}{y_{3}-y_{2}}\) (x – x_{1})

(y – y_{1}) (y_{3} – y_{2}) = -(x – x_{1}) (x_{3} – x_{2})

L_{1} = (x – x_{1})(x – x_{3}) + (y – y_{1})(y – y_{3}) = 0.

Similarly equations of the altitudes through B,C are

L_{2} = (x – x_{2}) (x_{3} – x_{1}) + (y – y_{2}) (y_{2} – y_{3}) = 0,

L_{3} = (x – x_{3}) (x_{1} – x_{2}) + (y – y_{3}) (y_{1} – y_{2}) = 0.

Now 1.L_{1} + 1.L_{2} + 1.L_{3} = 0

The altitudes L_{1} = 0, L_{2} = 0, L_{3} = 0 are concurrent.

Theorem:

The internal bisectors of the angles of a triangle are concurrent.

Theorem:

The perpendicular bisectors of the sides of a triangle are concurrent

**Point of Intersection of Two Straight Lines:**

Theorem:

The point of intersection of the two non parallel lines

a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0 is \(\left(\frac{b_{1} c_{2}-b_{2} c_{1}}{a_{1} b_{2}-a_{2} b_{1}}, \frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}\right)\)

Proof:

The lines are not parallel ^ Slopes are not equal

⇒ \(\frac{-a_{1}}{b_{1}} \neq \frac{-a_{2}}{b_{2}}\) ⇒ a_{1}b_{2} ≠ a_{2}b_{1} ⇒ a_{1}b_{2} – a_{2}b_{1} ≠ 0

Let P (α, β) be the point of intersection. Then a_{1} α + b_{1}β + C_{1} = 0 and a_{2} α + b_{2}β + c_{2} = 0.

Solving by the method of cross multiplication

Point of intersection is P = \(\left(\frac{b_{1} c_{2}-b_{2} c_{1}}{a_{1} b_{2}-a_{2} b_{1}}, \frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}\right)\)

Theorem:

The ratio in which the line L = ax + by + c = 0 divides the line segment joining A(x_{1}, y_{1}), B(x_{2}, y_{2}) is -L_{11} : L_{22}, where L_{11} = L(x_{1}, y_{1}) = ax_{1} + by_{1} + c and L_{22} = L(x_{2}, y_{2}) = ax_{2} + by_{2} + c.

Proof:

Let k : 1 be the ratio in which the line divides the line segment.

The point which divides in the ratio k : 1 is P = \(\left(\frac{k x_{2}+x_{1}}{k+1}, \frac{k y_{2}+y_{1}}{k+1}\right)\)

Since P lies on the line ax + by + c = 0 ⇒ a\(\left(\frac{k x_{2}+x_{1}}{k+1}\right)\) + b\(\left(\frac{k y_{2}+y_{1}}{k+1}\right)\) + c = 0

⇒ a(kx_{2} + x_{1}) + b(ky_{2} + y_{1}) + c(k + 1) = 0

⇒ k(ax_{2} + by_{2} + c) = -(ax_{1} + by_{1} + c)

⇒ k = –\(\frac{\left(a x_{1}+b y_{1}+c\right)}{a x_{2}+b y_{2}+c}\)

Required Ratio = -(ax_{1} + by_{1} + c) : (ax_{2} + by_{2} + c)

= -L_{11}: L_{22}.

Note:

The points A,B lie in the same side or opposite side of the line L = 0 according as L_{11}, L_{22} have the same sign or opposite signs.

- The points A,B are opposite sides of the line L = 0 iff L
_{11}and L_{22}have opposite signs. - The points A,B are same side of the line L = 0 iff L
_{11}and L_{22}have same sign.

**Concurrent Lines:**

Three or more lines are said to be concurrent if they have a point in common. The common point is called the point of concurrence.

**Condition for Concurrency of Three Straight Lines:**

Theorem:

The condition that the lines a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0, a_{3}x + b_{3}y + c_{3} = 0 to be concurrent is a_{3}(b_{1}c_{2} – b_{2}c_{1}) + b_{3}(c_{1}a_{2} – c_{2}a_{1}) + c_{3}(a_{1}b_{2} – a_{2}b_{1}) = 0.

Proof:

Suppose the given lines are concurrent.

Let P(α, β) be the point of concurrence.

Then a_{1} α + b_{1} β + c_{1} = 0 ………..(1)

a_{2} α + b_{2} β + c_{2}= 0 ………..(2)

a_{3} α + b_{3} β + c_{3}= 0 ………….(3)

By solving (1) and (2) we get

α = \(\frac{b_{1} c_{2}-b_{2} c_{1}}{a_{1} b_{2}-a_{2} b_{1}}\)

β = \(\frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}\)

Therefore P = \(\left(\frac{b_{1} c_{2}-b_{2} c_{1}}{a_{1} b_{2}-a_{2} b_{1}}, \frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}\right)\)

Substituting P in eq.(3), we get

a_{3}\(\left(\frac{b_{1} c_{2}-b_{2} c_{1}}{a_{1} b_{2}-a_{2} b_{1}}\right)\) + b_{3}\(\left(\frac{c_{1} a_{2}-c_{2} a_{1}}{a_{1} b_{2}-a_{2} b_{1}}\right)\) + c_{3} = 0

a_{3}(b_{1}c_{2} – b_{2}c_{1}) + b_{3}(c_{1}a_{2} – c_{2}a_{1}) + C3 (a_{1}b_{2} – a_{2}b_{1}) = 0 which is required Condition.

Above condition can be written in a determinant form as \(\left|\begin{array}{lll}

a_{1} & b_{1} & c_{1} \\

a_{2} & b_{2} & c_{2} \\

a_{3} & b_{3} & c_{3}

\end{array}\right|\) = 0

Problem:

Find the Condition that the lines ax + hy + g = 0, hx + by + f = 0, gx + fy + c = 0 to be Concurrent.

Answer:

Let P be the point of ConCurrenCe

aα + h β + g = 0 …………..(1)

hα + b β + f = 0 ………….(2)

gα + f β + c = 0 …………(3)

solving (1) and (2) we get

Sub these values in eq.(3)

g\(\left(\frac{h g-b g}{a b-h^{2}}\right)\) + f\(\left(\frac{g h-a f}{a b-h^{2}}\right)\) + c = 0

⇒ g(hf – bg) + f(gh – af) + C(ab – h^{2}) = 0

⇒ fgh – bg^{2} + fgh – af^{2} + abc – ch^{2} = 0

⇒ abc + 2fgh – af^{2} – bg^{2} – ch^{2} = 0

The Condition is abc + 2fgh – af^{2} – bg^{2} – ch^{2} = 0

**Family of Straight Lines – Concurrent Straight Lines:**

Theorem:

Suppose L_{1} = a_{1}x + b_{1}y + c_{1} = 0, L_{2} = a_{2}x + b_{2}y + c_{2} = 0 are two intersection lines.

(i) If (λ_{1}, λ_{2}) ≠ (0,0) then λ_{1}L_{1} + λ_{2}L_{2} = 0represents a straight line passing through the point of intersection of L_{1} = 0 and L_{2} = 0.

(ii) The equation of any line passing through the point of intersection of

L_{1} = 0, L_{2} = 0 is of the form where (λ_{1}, λ_{2}) ≠ (0,0).

Note: If L_{1} = 0, L_{2} = 0 are two intersecting lines, then the equation of any line other than L_{2} = 0 passing through their point of intersection can be taken as L_{1} + λL_{2} = 0 where λ is a parameter.

Theorem:

If there exists three constants p,q,r not all zero such that

p(a_{1}x + b_{1}y + c_{1}) + q(a_{2}x + b_{2}y + c_{2}) + r(a_{3}x + b_{3}y + c_{3}) = 0 for all x and y then the three lines a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0 and a_{3}x + b_{3}y + c_{3} = 0 in which no two of them are parallel, are concurrent.

Theorem:

Let L_{1} = a_{1}x + b_{1}y + c_{1} = 0, L_{2} = a_{2}x + b_{2}y + c_{2} = 0 represent two parallel lines.Then the straight line represented by λ_{1}L_{1} + λ_{2}L_{2} = 0 is parallel to each of the straight line L_{1} = 0 and L_{2} = 0.

**A Sufficient Condition for Concurrency of Three Straight Lines:**

Theorem:

If L_{1} = a_{1}x + b_{1}y + c_{1} = 0, L_{2} = a_{2}x + b_{2}y + c_{2} = 0, L_{3} = a_{3}x + b_{3}y + c_{3} = 0 are three straight lines, no two of which are parallel, and

if non-zero real numbers λ_{1}, λ_{2}, λ_{3} exist such that λ_{1} L_{1} + λ_{2} L_{2} + λ_{3} L_{3} = 0 then the straight lines L_{1} = 0, L_{2} = 0 and L_{3} = 0are concurrent.

**Length of The Perpendicular From A Point to A Straight Line and Distance Between Two Parallel Lines**

Theorem:

The perpendicular distance from a point P(x_{1}, y_{1}) to the line ax + by + c = 0 is \(\frac{\left|a x_{1}+b y_{1}+c\right|}{\sqrt{a^{2}+b^{2}}}\)

Proof:

Let the axes be translated to the point P(x_{1}, y_{1}).

Let (X,Y) be the new coordinates of (x, y). Then x = X + x_{1}, y = Y + y_{1}

The transformed equation of the given line is

a(X + x_{1}) + b(Y + y_{1}) + c = 0

⇒ aX + bY + (ax_{1} + by_{1} + c) = 0

The perpendicular distance from the new origin P to the line is (from normal form) The perpendicular distance from a point

P(x_{1}, y_{1}) to the line ax+ by + c = 0 is \(\frac{\left|a x_{1}+b y_{1}+c\right|}{\sqrt{a^{2}+b^{2}}}\)

**Distance Between Parallel Lines Theorem:**

The distance between the two parallel lines ax + by + c_{1} = 0 and ax + by + c_{2} = 0 is \(\frac{\left|c_{1}-c_{2}\right|}{\sqrt{a^{2}+b^{2}}}\)

Proof:

Given lines are ax + by + c_{1} = 0 …………(1)

ax + by + c_{2} = 0 …………(2)

Let P(x_{1}, y_{1}) be a point on the line (2).

Then

ax_{1} + by_{1} + c_{2} = 0

ax_{1} + by_{1} = -c_{2}.

Distance between the parallel lines = Perpendicular distance from P to line (i)

= \(\frac{\left|a x_{1}+b y_{1}+c_{1}\right|}{\sqrt{a^{2}+b^{2}}}=\frac{\left|c_{1}-c_{2}\right|}{\sqrt{a^{2}+b^{2}}}\)

**Foot of The Perpendicular:**

Theorem:

If (h, k) is the foot of the perpendicular from (x_{1}, y_{1}) to the line ax + by + c = 0 (a0, b0) then \(\frac{h-x_{1}}{a}=\frac{k-y_{1}}{b}=\frac{-\left(a x_{1}+b y_{1}+c\right)}{a^{2}+b^{2}}\).

Proof:

Let A = (x_{1}, y_{1}) P = (h, k)

P lies on ax + by + c = 0

ah + bk + c = 0

ah + bk = – c

Slope of \(\overline{A P}\) is \(\frac{k-y_{1}}{h-x_{1}}\)

Slope of given line is \(-\frac{a}{b}\)

\(\overline{A P}\) is perpendicular to the given line

⇒ \(\left(\frac{k-y_{1}}{h-x_{1}}\right)\left(-\frac{a}{b}\right)\) = -1

⇒ \(\frac{k-y_{1}}{b}=\frac{h-x_{1}}{a}\)

By the law of multipliers in ratio and proportion

**Image of A Point**

Theorem:

If (h, k) is the image of (x_{1}, y_{1}) w.r.t the line ax + by + c = 0 (a ≠ 0, b ≠ 0), then \(\frac{h-x_{1}}{a}=\frac{k-y_{1}}{b}=\frac{-2\left(x_{1}+b y_{1}+c\right)}{a^{2}+b^{2}}\).

Proof:

Let A(x_{1}, y_{1}), B(h, k)

Mid Point of is P = \(\left(\frac{x_{1}+h}{2}, \frac{y_{1}+k}{2}\right)\)

Since B is the image of A,therefore mid point P lies on ax + by + c = 0.

a\(\left(\frac{x_{1}+h}{2}\right)\) + b\(\left(\frac{y_{1}+k}{2}\right)\) + c = 0

⇒ ax_{1} + by_{1} + ah + bk + 2c = 0

⇒ ah + bk = -ax_{1} + by_{1} – 2c.

Slope of \(\overline{A B}\) is \(\frac{k-y_{1}}{h-x_{1}}\)

Slope of given line is \(-\frac{a}{b}\)

\(\overline{A B}\) is perpendicular to the given line a

⇒ \(\left(\frac{k-y_{1}}{h-x_{1}}\right)\left(-\frac{a}{b}\right)\) = -1

⇒ \(\frac{k-y_{1}}{b}=\frac{h-x_{1}}{a}\)

By the law of multipliers in ratio and proportion

Note :

- The image of (x
_{1}, y_{1}) w.r.t the line x = y is (y_{1}, x_{1}) - The image of (x
_{1}, y_{1}) w.r.t the line x + y = 0 is (-y_{1}, -x_{1})

Theorem:

If the four straight lines ax + by + p = 0, ax + by + q = 0, cx + dy + r = 0 and cx + dy + s = 0 form a parallelogram. Then the area of the parallelogram so formed is

\(\left|\frac{(p-q)(r-s)}{b c-a d}\right|\)

Proof:

Let L_{1} = ax + by + p = 0

L_{2} = ax + by + q = 0

L_{3} = cx + dy + r = 0

L_{4} = cx + dy + s = 0

Clearly

L_{1} ∥ L_{2} and L_{3} ∥ L_{4}. So L_{1} and L_{3} are nonparallel. Let be the angle between L_{1} and L_{3}.

Let d_{1} = distance between L_{1} and L_{2}= \(\frac{|p-q|}{\sqrt{a^{2}+b^{2}}}\)

Let d_{2} = distance between L_{3} and L_{4} = \(\frac{|r-s|}{\sqrt{c^{2}+d^{2}}}\)

Now cos θ = \(\frac{|a c+b d|}{\sqrt{\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)}}\) and sin θ = \(\sqrt{\frac{\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)-(a c+b d)^{2}}{\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)}}\)

= \(\frac{|b c-a d|}{\sqrt{\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)}}\)

Now area of the parallelogram is \(\frac{d_{1} d_{2}}{\sin \theta}=\left|\frac{(p-q))(r-s)}{b c-a d}\right|\)

**Angle Between Two Lines**

Theorem:

If θ is an angle between the lines a_{1}x + b_{1}y + c_{1} = 0, a_{2}x + b_{2}y + c_{2} = 0 then

cos θ = ±\(\frac{a_{1} a_{2}+b_{1} b_{2}}{\sqrt{a_{1}^{2}+b_{2}^{3}} \sqrt{a_{2}^{2}+b_{2}^{2}}}\)

Proof:

The lines passing through the origin and parallel to the given lines are

a_{1}x + b_{1}y = 0, ……… (1)

a_{2}x + b_{2}y = 0. …………….(2)

Let θ_{1}, θ_{2} be the inclinations of (1) and (2) respectively (θ1 > θ2)

Now θ is an angle between (1) and (2)

θ = θ_{1} – θ_{2}

P(-b_{1}, a_{1}) satisfies eq(1), the point lies on (1)

Similarly, Q(-b_{2}, a_{2}) lies on (2)

Let L and M be the projection of P, Q respectively on the x – axis.

Note :

- If is the acute angle between the lines then cos θ = \(\frac{\left|a_{1} a_{1}+b_{1} b_{2}\right|}{\sqrt{a_{1}^{2}+b_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}}}\)
- If is an angle between two lines, then is another angle between the lines.
- If is an angle between two lines are not a right angle then the angle between the lines means the acute angle between the lines.
- If 0 is an angle between the lines a
_{1}x + b_{1}y + c_{1}= 0, a_{2}x + b_{2}y + c_{2}= 0 then tan θ = \(\frac{a_{1} b_{2}-a_{2} b_{1}}{a_{1} a_{2}+b_{1} b_{2}}\) - If is the acute angle between the lines a
_{1}x + b_{1}y + c_{1}= 0, a_{2}x + b_{2}y + c_{2}= 0 then

tan θ = \(\left|\frac{a_{1} b_{2}-a_{2} b_{1}}{a_{1} a_{2}+b_{1} b_{2}}\right|=\left|\frac{a_{1} / b_{1}-a_{2} / b_{2}}{\left(a_{1} a_{2}\right) /\left(b_{1} b_{2}\right)+1}\right|\)

= \(\left|\frac{\left(-a_{1} / b_{1}\right)-\left(-a_{2} / b_{2}\right)}{1+\left(-a_{1} / b_{1}\right)\left(-a_{2} / b_{2}\right)}\right|\) where mj, m2 are the slopes of the lines.

Theorem:

The equation of the line parallel to ax + by + c = 0 and passing through (x_{1}, y_{1}) is a(x – x_{1}) + b(y – y_{1}) = 0.

Proof:

Slope of the given line is -a/b.

⇒ Slope of the required line is -a/b.(lines are parallel)

Equation of the required line is

y – y_{1} = –\(\frac{a}{b}\)(x – x_{1})

b(y – y_{1}) = -a(x – x_{1})

a(x – x_{1}) + b(y – y_{1}) = 0.

Note :

- The equation of a line parallel to ax + by + c = 0 may be taken as ax + by + k = 0.
- The equation of a line parallel to ax + by + c = 0 and passing through the origin is ax + by = 0.

Theorem:

The equation of the line perpendicular to ax + by + c = 0 and passing through (x_{1}, y_{1}) is b(x – x_{1}) – a(y – y_{1}) = 0.

Proof:

Slope of the given line is -a/b. ⇒ Slope of the required line is b/a. (since product of slopes = -1)

Equation of the required line is y – y_{1} = \(\frac{b}{a}\) (x – x_{1})

a(y – y_{1}) = b(x – x_{1})

b(x – x_{1}) – a(y – y_{1}) = 0.

Note :

- The equation of a line perpendicular to ax + by + c = 0 may be taken as bx – ay + k = 0
- The equation of a line perpendicular to ax + by + c = 0 and passing through the origin is bx – ay = 0.

**Concurrent Lines- Properties Related to A Triangle**

Theorem:

The medians of a triangle are concurrent.

Proof:

Let A(x_{1}, y_{1}), B(x_{2}, y_{2}), C(x_{3}, y_{3}) be the vertices of the triangle

Let D, E, F be the mid points of \(\overline{B C}, \overline{C A}, \overline{A B}\) respectively

Equation of \(\overline{A D}\)

y – y_{1} = (x – x_{1})

(y – y_{1}) (x_{2} + x_{3} – 2x_{1}) = (x – x_{1})(y_{2} + y_{3} – 2y_{1})

⇒ L_{1} ≡ (x – x_{1})(y_{2} + y_{3} – 2y_{1}) – (y – y_{1}) (x_{2} + x_{3} – 2x_{1}) = 0.

Similarly, the equations to \(\overline{B E}\) and \(\overline{C F}\) respectively are L_{2} ≡ (x – x_{2})(y_{3} + y_{1} – 2y_{2}) – (y – y_{2}) (x_{3} + x_{1} – 2x_{2}) = 0.

L_{3} ≡ (x – x_{3})(y_{1} + y_{2} – 2y_{3}) – (y – y_{3}) (x_{1} + x_{2} – 2x_{3}) = 0.

Now 1. L_{1} + 1.L_{2} + 1. L_{3} = 0

The medians L_{1} = 0, L_{2} =0, L_{3} = 0 are concurrent.

Theorem:

The altitudes of a triangle are concurrent.

Proof:

Let A(x_{1}, y_{1}), B(x_{2}, y_{2}), C(x_{3}, y_{3}) be the vertices of the triangle ABC.

Let AD, BE,CF be the altitudes.

Slope of \(\overline{B C}\) is \(\frac{y_{3}-y_{2}}{x_{3}-x_{2}}\) and AD ⊥ BC

Slope of the altitude through A is \(-\frac{x_{3}-x_{2}}{y_{3}-y_{2}}\)

Equation of the altitude through A is y – y_{1} = \(\frac{x_{3}-x_{2}}{y_{3}-y_{2}}\)(x – x_{1})

(y – y_{1}) (y_{3} – y_{2}) = -(x – x_{1}) (x_{3} – x_{2})

L_{1} = (x – x_{1})(x_{2} – x_{3}) + (y – y_{1})(y_{2} – y_{3}) = 0.

Similarly equations of the altitudes through B,C are

L_{2} = (x – x_{2}) (x_{3} – x_{1}) + (y – y_{2}) (y_{2} – y_{3}) = 0

L_{3} = (x – x_{3}) (x_{1} – x_{2}) + (y – y_{3}) (y_{1} – y_{2}) = 0.

Now 1.L_{1} + 1.L_{2} + 1.L_{3} = 0

The altitudes L_{1} = 0, L_{2} =0, L_{3} = 0 are concurrent.

Theorem:

The internal bisectors of the angles of a triangle are concurrent.

Theorem:

The perpendicular bisectors of the sides of a triangle are concurrent