Revision: Coordinate Geometry >> Straight Lines Applied 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 = \[\left|\frac{c_{1}-c_{2}}{\sqrt{a^{2}+b^{2}}}\right|\]

The vertices of the rectangle ABCD are A(2, -1), B(5, -1), C(5, 6) and D(2, 6) Now,

`"Coordinates of midpoint of" AC = ((2+5)/2 , (-1+6)/2) = (7/5 ,5/2)`

`"Coordinates of midpoint of " BD = ((5+2)/2 , (-1+6)/2)= (7/2,5/2)`

Since, the midpoints of AC and BD coincide, therefore the diagonals of rectangle ABCD bisect each other.

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

As per the question, we have

`sqrt((x-a-b)^2 +(y-b+a)^2 ) = sqrt((x-a+b)^2 +(y-a-b)^2)`

`⇒(x-a-b)^2 +(y-b+a)^2 = (x-a+b)^2 +(y-a-b)^2`      (Squaring both sides) 

`⇒x^2 + (a+b)^2 -2x (a+b) +y^2 +(a-b)^2 -2y(a-b)=x^2 +(a-b)^2 -2x(a-b)+y^2 +(a+b)^2 -2y (a+b)`

`⇒-x(a+b) - y (a-b) = -x(a-b) -y(a+b)`

`⇒-xa -xb -ay +by = -xa + bx -ya-by`

⇒ by=bx

Hence, . bx = ay 

Let A (-4,-1); B (-2,-4); C (4, 0) and D (2, 3) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a rectangle.

So we should find the lengths of opposite sides of quadrilateral ABCD.

`AB = sqrt((-2+4)^2) + (-4 + 1)^2)`

`= sqrt(4 + 9)`

`= sqrt13`

`CD = sqrt((4 - 2)^2 + (0 - 3)^2)`

`= sqrt(4 +9)`

`= sqrt13`

Opposite sides are equal. So now we will check the lengths of the diagonals.

`AC = sqrt((4 + 4)^2 + (0 + 1)^2)`

`= sqrt(64 + 1)`

`= sqrt(65)`

`BD = sqrt((2 + 2)^2 + (3 + 4)^2)`

`= sqrt(16 + 49)`

`= sqrt65`

Opposite sides are equal as well as the diagonals are equal. Hence ABCD is a rectangle.

The given points are  A (-4,-1); B (-2,-4); C (4, 0) and D (2, 3) .

`AB = sqrt({-2-(-4)}^2 + { -4-(-1)}^2) = sqrt ((2)^2+(-3)^2) = sqrt(4+9) = sqrt(13) ` units

` BC = sqrt({ 4-(-2)}^2+{0-(-4)}^2) = sqrt((6)^2 +(4)^2) = sqrt(36+16) = sqrt(52) = 2 sqrt(13)  units`

`CD = sqrt((2-4)^2 +(3-0)^2) = sqrt((-2)^2 +(3)^2) = sqrt(4+9) = sqrt(13)  units`

`AD = sqrt({2-(-4)}^2 + {3-(-1)}^2) = sqrt((6)^2 +(4)^2) = sqrt(36+16) = sqrt(52) = 2 sqrt(13) units`

`Thus , AB = CD = sqrt(13)   units and BC = AD = 2 sqrt(13)   units`

Also , `AC = sqrt({4-(-4)}^2+{0-(-1)}^2) = sqrt ((8)^2+(1)^2 ) = sqrt(64+1) = sqrt(65)  units`

`BD = sqrt({2-(-2)}^2 +{3-(-4)}^2) = sqrt((4)^2 +(7)^2) = sqrt(16+49) = sqrt(65)  units`

Also, diagonal AC = diagonal BD

Hence, the given points form a rectanglr

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”.