Revision: Coordinate Geometry >> Straight Lines Mathematics ISC (Commerce) Class 11 CISCE

The slope m of a line is m = tan⁡θ

where θ is the inclination of the line with the positive x-axis.

An equation of the form ax + by + c = 0 represents a straight line and is known as a linear equation.

The equation of the line with intercepts a and b is \[\frac{x}{a} + \frac{y}{b} = 1\] .Since the line meets the coordinate axes at A and B, the coordinates are A (a, 0) and B (0, b).
Let the given point be P (3, 4).
Here,

\[AP : BP = 2 : 3\]

\[\therefore 3 = \frac{2 \times 0 + 3 \times a}{2 + 3}, 4 = \frac{2 \times b + 3 \times 0}{2 + 3}\]

\[ \Rightarrow 3a = 15, 2b = 20\]

\[ \Rightarrow a = 5, b = 10\]

Hence, the equation of the line is

\[\frac{x}{5} + \frac{y}{10} = 1\]

\[ \Rightarrow 2x + y = 10\]

If L₁: a₁x + b₁y + c₁ = 0 and L₂: a₂x + b₂y + c₂ = 0 represent two intersecting lines, then equation L₁ + λL₂ = 0, λ ∈ R, represents a family of lines.

Three or more lines are concurrent if they meet at a single point.

The two mutually perpendicular number lines intersecting each other at their zeroes are called rectangular axes or coordinate axes, or axes of reference. 

The position of a point in a plane is expressed by a pair of numbers, one concerning the x-axis and the other concerning the y-axis. called co-ordinates. 

• x → distance from y-axis (abscissa)

• y → distance from x-axis (ordinate)

The equation of the locus of a point is the algebraic relation which is satisfied by the coordinates of every point on the locus of the point.

Locus is the path traced by a moving point, which moves so as to satisfy a certain given condition/conditions. 

\[m=\frac{y_2-y_1}{x_2-x_1}\]

From general form:

• Slope (m) = −a / b
• Y-intercept = −c / b

If ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of parallel straight lines, then the distance between them is given by

\[2\sqrt{\frac{g^{2}-ac}{a(a+b)}}\mathrm{or}2\sqrt{\frac{f^{2}-bc}{b(a+b)}}\]

For point (x₁, y₁) and line ax + by + c = 0,

\[p=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\]

For lines ax + by + c₁ = 0 and ax + by + c₂ = 0,

P = \[\left|\frac{c_{1}-c_{2}}{\sqrt{a^{2}+b^{2}}}\right|\]

As per the question, we have

AP = BP

`⇒ sqrt((x -5)^2 +(y-1)^2) = sqrt((x+1)^2 +(y-5)^2)`

`⇒(x-5)^2 +(y-1)^2 = (x+1)^2 +(y-5)^2`          (Squaring both sides) 

`⇒x^2 - 10x +25 + y^2 -2y +1 = x^2 +2x +1+y^2 -10y+25`

⇒ –10x – 2y = 2x – 10y

⇒ 8y = 12x

⇒ 3x = 2y

Nature of Slope

• m > 0 → rising line

• m < 0 → falling line

• m = 0 → horizontal line

• m = ∞→ vertical line

Parallel Lines 

Two lines are parallel ⇔ , their slopes are equal, m₁ = m₂

​Perpendicular Lines

Two lines are perpendicular ⇔ m₁ × m₂ = −1

​Collinearity of Three Points

Points A, B, and C are collinear

Method 1: Distance method

AB + BC = AC

Method 2: Slope method

Slope of AB = Slope of BC

Position of a Point:

For line: ax₁ + by₁ + c

• If ax₁ + by₁ + c = 0 → Point lies on the line
• If ax₁ + by₁ + c < 0 → Point lies on one side (origin side)
• If ax₁ + by₁ + c > 0 → Point lies on other side

Condition for concurrency:

For lines
a₁x + b₁y + c₁ = 0
a₂x + b₂y + c₂ = 0
a₃x + b₃y + c₃ = 0

\[\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}=0\]

Sign Convention

• Right of y-axis → +x

• Left of y-axis → −x

• Above x-axis → +y

• Below x-axis → −y

Standard Line Results

• x = 0 → y-axis

• y = 0 → x-axis

• x = a → line parallel to the y-axis

• y = b → line parallel to the x-axis

Quadrant Reminder

• Step I: Take any point P(x, y) on the locus.
• Step II: Write down the geometrical condition of the locus.
• Step III: Convert the geometrical condition into an algebraic equation involving x and y.
• Step IV: Simplify the equation to get the required “equation of the locus”.