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Revision: Co-ordinate Geometry JEE Main Co-ordinate Geometry

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Definitions [16]

Definition: Equation of Locus

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.

Definition: Locus

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

Definition: Slope

The slope m of a line is m = tan⁡θ

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

Definition: Intercepts on Axes

x-intercept: Point where a line cuts the x-axis, y = 0

y-intercept: Point where a line cuts the y-axis, x = 0

Definition: Linear Equation

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

Find the equation of the line which passes through the point (3, 4) and is such that the portion of it intercepted between the axes is divided by the point in the ratio 2:3.

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 (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

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

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

Definition: Conic Sections

A conic section is the locus of a point such that the ratio of its distance from a fixed point (focus) to a fixed line (directrix) is constant.

Definition: Axis

The straight line passing through the focus and perpendicular to the directrix is called the axis of the conic section.

Definition: Vertex

The points of intersection of the conic section and the axis are called the vertices of the conic section.

Definition: Latusrectum

The chord passing through the focus and perpendicular to the axis is called the latus rectum of the conic section.

Definition: Focal Chord

A chord of a conic passing through the focus is called a focal chord.

Definition: Double Ordinate

A straight line drawn perpendicular to the axis and terminating at both ends of the curve is a double ordinate of the conic section.

Definition: Centre

The point which bisects every chord of the conic passing through it is called the centre of the conic section.

Definition: Parabola

A parabola is the locus of a point which moves in a plane such that its distance from a fixed point (i.e. focus) is always equal to its distance from a fixed straight line (i.e. directrix).

Definition: Hyperbola

A hyperbola is the locus of a point in a plane which moves in such a way that the ratio of its distance from a fixed point (i.e. focus) to its distance from a fixed line (i.e. directrix) is always constant and greater than unity.

Definition: Ellipse

An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is constant.

Formulae [14]

Formula: Distance Formula

The distance between P(x1, y1) and Q(x2, y2) is

\[\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\]

 The distance of a point P(x, y) from the origin is

\[\sqrt{x^2+y^2}\]

Formula: Section Formula

\[P\left(\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}\right)\]

Formula: Slope Between Two Points

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

Formula: Intercepts on Axes

For line ax + by + c = 0

x-intercept:

\[\left(-\frac{c}{a},0\right)\]

y-intercept:

\[\left(0,-\frac{c}{b}\right)\]

Formula: Distance of a Point from a Line

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

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

Formula: Distance Between A Pair of Parallel Straight Lines

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

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

Formula: Distance Between Two Parallel Lines

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

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

Formula: Slope & Intercept

From general form:

  • Slope (m) = −a / b
  • Y-intercept = −c / b
Formula: Eccentricity (e)

$$e = \frac{\text{distance from focus}}{\text{distance from directrix}}$$

Formula: Distance between Parallel Lines

\[SD=\left|\frac{\left(a_{2}-a_{1}\right)\times b}{\left|b\right|}\right|\]

Formula: Distance between Skew Lines

Vector Form:

\[\mathbf{d}=\left|\frac{(\overline{\mathbf{b}}_{1}\times\overline{\mathbf{b}}_{2}).(\overline{\mathbf{a}}_{2}-\overline{\mathbf{a}}_{1})}{\left|\overline{\mathbf{b}}_{1}\times\overline{\mathbf{b}}_{2}\right|}\right|\]

Cartesian Form:

\[\mathbf{d}=\left|\frac{ \begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ \mathbf{a}_1 & \mathbf{b}_1 & \mathbf{c}_1 \\ \mathbf{a}_2 & \mathbf{b}_2 & \mathbf{c}_2 \end{vmatrix}}{\sqrt{\left(\mathbf{a}_1\mathbf{b}_2-\mathbf{a}_2\mathbf{b}_1\right)^2+\left(\mathbf{a}_1\mathbf{c}_2-\mathbf{a}_2\mathbf{c}_1\right)^2+\left(\mathbf{b}_1\mathbf{c}_2-\mathbf{b}_2\mathbf{c}_1\right)^2}}\right|\]

Formula: Distance Between A Pair of Parallel Straight Lines

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

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

Formula: Distance of a Point from a Line

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

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

Formula: Distance Between Two Parallel Lines

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

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

Theorems and Laws [5]

If the point (x, y) is equidistant from the points (a + b, b – a) and (a – b, a + b), prove that bx = ay.

Let P(x, y), Q(a + b, b – a) and R (a – b, a + b) be the given points. Then

PQ = PR   ...(Given)

⇒ `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`

⇒ x2 – 2x(a + b) + (a + b)2 + y2 – 2y(b – a) + (b – a)2 = x2 + (a – b)2 – 2x(a – b) + y2 – 2(a + b) + (a + b)2

⇒ –2x(a + b) – 2y(b – a) = –2x(a – b) – 2y(a + b)

⇒ ax + bx + by – ay = ax – bx + ay + by

⇒ 2bx = 2ay

⇒ bx = ay

Let `A(bara)` and `B(barb)` be any two points in the space and `R(barr)` be the third point on the line AB dividing the segment AB externally in the ratio m : n, then prove that `barr = (mbarb - nbara)/(m - n)`.


As the point R divides the line segment AB externally, we have either A-B-R or R-A-B.

Assume that A-B-R and `bar(AR) : bar(BR)` = m : n

∴ `(AR)/(BR) = m/n` so n(AR) = m(BR) 

As `n(bar(AR))` and `m(bar(BR))` have same magnitude and direction,

∴ `n(bar(AR)) = m(bar(BR))`

∴ `n(barr - bara) = m(barr - barb)`

∴ `nbarr - nbara = mbarr - mbarb`

∴ `mbarr - nbarr = mbarb - nbara`

∴ `(m - n)barr = mbarb - nbara`

∴ `barr = (mbarb - nbara)/(m - n)`

Hence proved.

Let `A(bara)` and `B(barb)` are any two points in the space and `R(barr)` be a point on the line segment AB dividing it internally in the ratio m : n, then prove that `barr = (mbarb + nbara)/(m + n)`.

R is a point on the line segment AB(A – R – B) and `bar(AR)` and `bar(RB)` are in the same direction.

Point R divides AB internally in the ratio m : n

∴ `(AR)/(RB) = m/n`

∴ n(AR) = m(RB)

As `n(bar(AR))` and `m(bar(RB))` have same direction and magnitude,

`n(bar(AR)) = m(bar(RB))`

∴ `n(bar(OR) - bar(OA)) = m(bar(OB) - bar(OR))`

∴ `n(vecr - veca) = m(vecb - vecr)`

∴ `nvecr - nveca = mvecb - mvecr`

∴ `mvecr + nvecr = mvecb + nveca`

∴ `(m + n)vecr = mvecb + nveca`

∴ `vecr = (mvecb + nveca)/(m + n)`

By vector method prove that the medians of a triangle are concurrent.


Let A, B and C be vertices of a triangle.

Let D, E and F be the mid-points of the sides BC, AC and AB respectively.

Let `bara, barb, barc, bard, bare` and `barf` be position vectors of points A, B, C, D, E and F respectively.

Therefore, by mid-point formula,

∴ `bard = (barb + barc)/2, bare = (bara + barc)/2` and `barf = (bara + barb)/2`

∴ `2bard = barb + barc, 2bare = bara + barc` and `2barf = bara + barb`

∴ `2bard + bara = bara + barb + barc`, similarly `2bare + barb = 2barf + barc = bara + barb + barc`

∴ `(2bard + bara)/3 = (2bare + barb)/3 = (2barf + barc)/3 = (bara + barb + barc)/3 = barg`  ...(Say)

Then we have `barg = (bara + barb + barc)/3 = ((2)bard + (1)bara)/(2 + 1) = ((2)bare + (1)barb)/(2 + 1) = ((2)barf + (1)barc)/(2 + 1)`

If G is the point whose position vector is `barg`, then from the above equation it is clear that the point G lies on the medians AD, BE, CF and it divides each of the medians AD, BE, CF internally in the ratio 2 : 1.

Therefore, three medians are concurrent.

If D, E, F are the midpoints of the sides BC, CA, AB of a triangle ABC, prove that `bar(AD) + bar(BE) + bar(CF) = bar0`.

Let `bara, barb, barc, bard, bare, barf` be the position vectors of the points A, B, C, D, E, F respectively.

Since D, E, F are the midpoints of BC, CA, AB respectively, by the midpoint formula

`bard = (barb + barc)/2, bare = (barc + bara)/2, barf = (bara + barb)/2`

∴ `bar(AD) + bar(BE) + bar(CF) = (bard - bara) + (bare - barb) + (barf - barc)`

= `((barb + barc)/2 - bara) + ((barc + bara)/2 - barb) + ((bara + barb)/2 - barc)`

= `1/2barb + 1/2barc - bara + 1/2barc + 1/2bara - barb + 1/2bara + 1/2barb - barc`

= `1/2(barb + barc - 2bara + bar c + bara - 2barb + bara + barb - 2barc)`

= `(bara + barb + barc) - (bara + barb + barc) = bar0`. 

Key Points

Key Points: Locus
  • 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”.
Key Points: Concept of Slope

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 ⇔

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

Key Points: Intercepts on Axes
  • x-intercept:
    Right of origin → positive
    Left of origin → negative

  • y-intercept:
    Above origin → positive
    Below origin → negative

Key Points: Equations of Line in Different Forms
Form Formula
X-axis y = 0
Y-axis x = 0
Parallel to the X-axis y = b or y = -b
Parallel to the Y-axis x = a or x = -a
Slope-point form y − y₁ = m(x − x₁)
Two-point form \[\frac{y-y_{1}}{y_{1}-y_{2}}=\frac{x-x_{1}}{x_{1}-x_{2}}\]
Slope-intercept form y = mx + c
Intercept form \[\frac{x}{\mathrm{a}}+\frac{y}{\mathrm{b}}=1\]
Normal form x cosα + y sinα = p
Parametric form \[\frac{x-x_{1}}{\cos\theta}=\frac{y-y_{1}}{\sin\theta}=r\]

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
Key Points: Equation of Tangent and Condition of Tangency

For Standard Circle: x² + y² = a²

Sr. No. Description Formula
i. Tangent at a point (x₁, y₁) xx₁ + yy₁ = a²
ii. Parametric form of tangent at P(θ) x cosθ + y sinθ = a
iii. Condition of tangency for the line y = mx + c \[\mathrm{c=\pm a~\sqrt{1+m^{2}}}\]
  Point of contact \[\left(\frac{-\mathrm{a}^{2}\mathrm{m}}{\mathrm{c}},\frac{\mathrm{a}^{2}}{\mathrm{c}}\right)\]
iv. Equation of tangent in terms of its slope m \[y=\mathrm{m}x\pm\mathrm{a}\sqrt{1+\mathrm{m}^{2}}\]
v. Length of tangent from the point (x₁, y₁) \[\sqrt{S_{1}}=\sqrt{x_{1}^{2}+y_{1}^{2}-a^{2}}\]
vi. Equation of the Director circle x² + y² = 2a²

For General Circle: x² + y² + 2gx + 2fy + c = 0

Sr. No. Description Formula
i. Tangent at a point (x₁, y₁) xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
ii. Length of tangent from the point (x₁, y₁) \[\sqrt{S_{1}}=\sqrt{x_{1}^{2}+y_{1}^{2}+2gx_{1}+2fy_{1}+c}\]

Number of Common Tangents:

Case Diagram No. of Tangents Condition
Disjoint circles 4 d > r₁ + r₂
Touch externally 3 d = r₁ + r₂
Intersecting circles 2 d < r₁ + r₂
Touch internally 1 d = \[\left|\mathbf{R}_{1}-\mathbf{R}_{2}\right|\]
Concentric circles 0 d = 0

Equation of a pair of tangents:

(x² + y² − a²)(x₁² + y₁² − a²) = (xx₁ + yy₁ − a²)²

Key Points: Parabola
Property y² = 4ax y² = −4ax x² = 4ay x² = −4ay
Vertex (0, 0) (0, 0) (0, 0) (0, 0)
Focus (a, 0) (−a, 0) (0, a) (0, −a)
Directrix x + a = 0 x − a = 0 y + a = 0 y − a = 0
Axis y = 0 y = 0 x = 0 x = 0
Axis of Symmetry X-axis X-axis Y-axis Y-axis
Eccentricity 1 1 1 1
Latus Rectum Length 4a 4a 4a 4a
Endpoints of Latus Rectum (a, ±2a) (−a, ±2a) (±2a, a) (±2a, −a)
Equation of Latus Rectum x = a x = −a y = a y = −a
Tangent at Vertex x = 0 x = 0 y = 0 y = 0
Parametric Equations x = at², y = 2at x = −at², y = 2at x = 2at, y = at² x = 2at, y = −at²
Parametric Point (at², 2at) (−at², 2at) (2at, at²) (2at, −at²)
Focal Distance of P(x₁,y₁) x₁ + a a − x₁ y₁ + a a − y₁
Key Points: Hyperbola
Property

Standard Hyperbola

\[\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\]

Conjugate Hyperbola

\[\frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1\]

Centre (0, 0) (0, 0)
Vertices (±a, 0) (0, ±b)
Transverse Axis Length 2a 2b
Conjugate Axis Length 2b 2a
Foci (±ae, 0) (0, ±be)
Eccentricity \[\mathrm{e}=\frac{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}{\mathrm{a}}\] \[\mathrm{e}=\frac{\sqrt{\mathrm{b}^2+\mathrm{a}^2}}{\mathrm{b}}\]
Relation \[\mathbf{b}^2=\mathbf{a}^2(\mathbf{e}^2-1)\] \[\mathbf{a}^2=\mathbf{b}^2(\mathbf{e}^2-1)\]
Directrices \[x=\pm\frac{\mathrm{a}}{\mathrm{e}}\] \[y=\pm\frac{\mathrm{b}}{\mathrm{e}}\]
Length of Latus Rectum \[\frac{2\mathrm{b}^2}{\mathrm{a}}\] \[\frac{2\mathrm{a}^2}{\mathrm{b}}\]
Ends of Latus Rectum \[\left(\pm ae,\pm\frac{b^{2}}{a}\right)\] \[\left(\pm\frac{a^{2}}{b},\pm e\right)\]
Distance between Foci 2ae 2be
Difference of Focal Radii 2a 2b
Axis Equations Transverse: y = 0, Conjugate: x = 0 Transverse: x = 0, Conjugate: y = 0
Parametric Equations x = a secθ, y = b tanθ x = a tanθ, y = b secθ
Parametric Point (a secθ, b tanθ) (a tanθ, b secθ)
Tangent at Vertex x = ±a y = ±b
Key Points: Ellipse and its Types
Fundamental Terms Horizontal Ellipse (a>b) Vertical Ellipse (a<b)
Equation \[\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\] \[\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\]
Centre (0,0) (0,0)
Vertices (±a,0) (0,±b)
Length of major axis 2a 2b
Length of minor axis 2b 2a
Foci (±ae,0) (0, ±be)
Relation between (a,b,e) \[\mathrm{b}^{2}=\mathrm{a}^{2}(1-\mathrm{e}^{2})\] \[\mathbf{a}^{2}=\mathbf{b}^{2}(1-\mathbf{e}^{2})\]
Eccentricity \[\mathrm{e}=\frac{\sqrt{\mathrm{a}^{2}-\mathrm{b}^{2}}}{\mathrm{a}}\] \[\mathrm{e}=\frac{\sqrt{\mathrm{b}^{2}-\mathrm{a}^{2}}}{\mathrm{b}}\]
Equation of directrices \[x=\pm\frac{\mathrm{a}}{\mathrm{e}}\] \[y=\pm\frac{b}{e}\]
Distance between foci 2ae 2be
Distance between directrices \[\frac{2a}{e}\] \[\frac{2b}{e}\]
Length of latus rectum \[\frac{2\mathrm{b}^2}{a}\] \[\frac{2\mathrm{a}^2}{b}\]
Endpoints of the latus rectum \[\left(\pm ae,\pm\frac{b^{2}}{a}\right)\] \[\left(\pm\frac{a^{2}}{b},\pm be\right)\]
Equation of axes Major: (y = 0), Minor: (x = 0) Major: (x = 0), Minor: (y = 0)
Parametric equations \[\begin{cases} x=a\cos\alpha \\ y=b\sin\alpha & \end{cases}\] \[\begin{cases} x=a\cos\alpha \\ y=b\sin\alpha & \end{cases}\]
Focal distances \[\mid SP\mid=\left(a-ex_{1}\right)\mathrm{and}\mid S^{\prime}P\mid=\left(a+ex_{1}\right)\] \[\mid SP\mid=(b-ey_{1})\mathrm{~and}\mid S^{\prime}P\mid=(b+ey_{1})\]
Sum of focal radii 2a 2b
Equation of the tangent at the vertex (x = ± a) (y = ± b)

Concepts [38]

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