Option 2 : 4/3

__Concept:__

- Slope intercept form of line:
**y = mx + c**, Where ‘m’ is the slope of the line and ‘c’ is the y-intercept. - The "point-slope" form of the equation of a straight line is:
**y - y**_{1}= m(x - x_{1}) - When two lines are perpendicular, the product of their slope is -1.
- If m is the slope of a line, then the slope of a line perpendicular to it is -1/m.

__Calculation:__

Given: 3x + y = 3

⇒ y = -3x + 3

The slope of line = m = -3

Then the slope of a line perpendicular to it is -1/m = 1/3

The equation of line passing through (2, 2) with slope "1/3" is

y - 2 = (1/3) (x - 2)

⇒ 3y - 6 = x - 2

⇒ 3y = x + 4

\(\Rightarrow {\rm{y}} = {\rm{\;}}\frac{1}{3}{\rm{x}} + {\rm{\;}}\frac{4}{3}\)

So, the y-intercept of line is 4/3

Option 2 : \(\frac 5 2\)

**Concept:**

If three points (x1, y1), (x2, y2) and (x3, y3) are** collinear** then the area of the triangle determined by the three points is zero.

\(\left| {\begin{array}{*{20}{c}} {{{\rm{x}}_1}}&{{{\rm{y}}_1}}&1\\ {{{\rm{x}}_2}}&{{{\rm{y}}_2}}&1\\ {{{\rm{x}}_3}}&{{{\rm{y}}_3}}&1 \end{array}} \right|{\rm{\;}} = {\rm{\;}}0\)

or

If three or more points are **collinear** then the slope of any two pairs of points is same.

For example, let three points A, B and C are collinear then

slope of AB = slope of BC = slope of AC

Slope of the line if two points \(({{\rm{x}}_1},{{\rm{y}}_1}){\rm{\;and\;}}\left( {{{\rm{x}}_2},{\rm{\;}}{{\rm{y}}_2}} \right)\) are given by:

\(\Rightarrow \left( {\bf{m}} \right) = \;\frac{{{{\bf{y}}_2} - {{\bf{y}}_1}}}{{{{\bf{x}}_2} - {{\bf{x}}_1}}}{\rm{\;}}\)

**Calculation:**

Given: the points (-2, -5), (2, -2) and (8, a) are collinear

\(\begin{vmatrix} -2 &-5 &1 \\ 2 &-2 &1 \\ 8& \rm a &1 \end{vmatrix} = 0 \\\Rightarrow -2{(-2-\rm a)}-(-5)(2-8)+1(2a+16)=0\\\Rightarrow4+2\rm a-30+2\rm a + 16 = 0\\\Rightarrow 4\rm a-10=0\\\Rightarrow 4\rm a = 10\\\therefore \rm a = \frac{5}{2}\)

Option 2 : 3/10 unit

__Concept:__

**Distance between parallel lines:**

- The distance between the lines y = mx + c
_{1}and y = mx + c_{2}is \(\frac{{\left| {{{\rm{c}}_1} - {{\rm{c}}_2}} \right|}}{{\sqrt {1{\rm{\;}} + {\rm{\;}}{{\rm{m}}^2}} }}\) - The distance between the lines ax + by + c
_{1}= 0 and ax + by + c_{2}= 0 is \(\frac{{\left| {{{\rm{c}}_1} - {{\rm{c}}_2}} \right|}}{{\sqrt {{{\rm{a}}^2} + {{\rm{b}}^2}} }}\)

__Calculation:__

Given lines are 6x + 8y + 15 = 0 and 3x + 4y + 9 = 0

⇒ 6x + 8y + 15 = 0

Take 2 common from above equation, we get

⇒ 3x + 4y + 15/2 = 0 ---(1)

And 3x + 4y + 9 = 0 ---(2)

Equation 1 and 2 are parallel to each other.

∴ The distance between the lines = \(\frac{{\left| {\frac{{15}}{2}{\rm{\;}} - {\rm{\;}}9} \right|}}{{\sqrt {{3^2} + {4^2}} }} = \frac{{\left( {\frac{3}{2}} \right)}}{5} = \frac{3}{{10}}\)

Option 3 : \(\rm 3x + 2y - \frac {23}{5} = 0\)

__Concept:__

If the two lines are **parallel**, then their **slopes will be equal**.

The **"point-slope"** form of the equation of a straight line is: y – y1 = m (x – x1)

Here (x1, y1) is point on the line and m is the slope of line.

**Calculation:**

Given lines are x - 2y = 1 and x + 3y = 2

x - 2y = 1 .... (1)

x + 3y = 2 ..... (2)

equation (2) - equation (1), we get

⇒ (x + 3y) - (x - 2y) = 2 - 1

⇒ 5y = 1

∴ y = 1/5

Put the value of y in equation (1), we get

x = 7/5

Point of intersection: (x, y) = (x_{1}, y_{1}) = \((\frac{7}{5},\frac{1}{5})\)

Let slope of the straight line 3x + 2y = 0 is m_{1},

So, slope = m_{1} = \(\frac{-3}{2}\)

We know that when two lines are parallel, then their slopes will be equal.

∴ slope = m = m_{1} = \(\frac{-3}{2}\)

Now equation of the line is y – y_{1} = m (x – x_{1})

\(\rm \Rightarrow (y - \frac{1}{5}) = \frac{-3}{2} (x - \frac{7}{5})\\ \Rightarrow \frac{(5y -1)}{5} = \frac{-3}{2}\frac{(5x-7)}{5}\\ \Rightarrow 2 \times(5y -1) =-3\times(5x-7)\\ \Rightarrow 10y - 2 = -15x + 21\\\Rightarrow 15x+10y - 23 =0 \\\Rightarrow 5(3x+2y - \frac{23}{5}) =0 \\\therefore 3x+2y - \frac{23}{5} =0\)

Option 3 : (0, 2√3) or (3, -√3)

__Concept:__

Let A (x_{1}, y_{1}) and B (x_{2}, y_{2})

- \({\rm{Distance\;between\;A\;and\;B\;}} = \sqrt {{\rm{\;}}{{\left( {{{\rm{x}}_2} - {\rm{\;}}{{\rm{x}}_1}} \right)}^2} + {\rm{\;}}{{\left( {{{\rm{y}}_2} - {\rm{\;}}{{\rm{y}}_1}} \right)}^2}} \)

__Calculation:__

Given:

Two vertices of an equilateral triangle are (0, 0) and (3, √3).

Let the third vertex of the equilateral triangle be (x, y)

Distance between (0, 0) and (x, y) = Distance between (0, 0) and (3, √3) = Distance between (x, y) and (3, √3)

\(\sqrt {{{\left( {x - 0} \right)}^2} + \;{{\left( {y - 0} \right)}^2}} = \;\sqrt {{{\left( {3 - 0} \right)}^2} + \;{{\left( {\sqrt 3 - 0} \right)}^2}} = \;\sqrt {{{\left( {x - 3} \right)}^2} + \;{{\left( {y - \sqrt 3 } \right)}^2}} \)

\(\Rightarrow \sqrt {{x^2} + \;{y^2}} = \;\sqrt {12} = \;\sqrt {{{\left( {x - 3} \right)}^2} + \;{{\left( {y - \sqrt 3 } \right)}^2}} \)

Squaring both sides, we get

⇒ x^{2} + y^{2} = 12 = (x – 3)^{2} + (y - √3)^{2}

Now,

x^{2} + y^{2} = 12 …. (1)

x^{2} + 9 - 6x + y^{2} + 3 - 2√3y = 12 …. (2)

Solving equations 1 and 2,

x^{2} + 9 - 6x + y^{2} + 3 - 2√3y = x^{2} + y^{2}

⇒ - 6x - 2√3y + 12 = 0

⇒ 3x + √3y = 6

⇒ x = (6 - √3y) / 3

Now, substituting the value of x in equation 1, we get

\(\Rightarrow \;{\left( {\frac{{6 - \sqrt 3 y}}{3}} \right)^2} + \;{y^2} = 12\)

\(\Rightarrow \frac{{36 + 3{y^2} - 12\sqrt 3 y}}{9} + \;{y^2} = 12\)

⇒ 36 + 3y^{2} - 12√3y + 9y^{2} = 108

⇒ - 12√3y + 12y^{2} - 72 = 0

⇒ y2 - √3y - 6 = 0

⇒ (y - 2√3) (y + √3) = 0

⇒ y = 2√3 or - √3

If y = 2√3, x = (6 - 6) / 3 = 0

If y = -√3, x = (6 + 3) / 3 = 3

So, the third vertex of the equilateral triangle = (0, 2√3) or (3, -√3).

Option 1 : isosceles

** Concept**:

I. Any triangle in which all the sides are of different length is called a **scalene** triangle.

II. Any triangle in which length of any two sides of the triangle is same is said to be an **isosceles** triangle.

III. Any triangle in which all the sides of the triangle are of equal length is said to be an **equilateral** triangle.

** Calculation**:

Given: The straight lines x + y – 4 = 0, 3x + y – 4 = 0 and x + 3y – 4 = 0 form a triangle.

Let, A be the point of intersection of line x + y – 4 = 0 and 3x + y – 4 = 0. So, by solving the two equations we get the co-ordinate of point A which is (0, 4)

Similarly, let B be the point of intersection of line 3x + y – 4 = 0 and x + 3y – 4 = 0. So, by solving the two equations we get the co-ordinate of point B which is (1, 1).

Similarly, let C be the point of intersection of line x + y – 4 = 0 and x + 3y – 4 = 0. So, by solving the two equations we get the co-ordinate of point C which is (4, 0).

So, Length of side AB \( = \sqrt {{1^2} + {3^2}} = \sqrt {10}\)

Similarly, Length of side BC \(= \sqrt {{3^2} + {1^2}} = \sqrt {10}\)

Similarly, Length of side AC \(= \sqrt {{4^2} + {4^2}} = 4\sqrt 2\)

So, as AB = BC ≠ AC.

Hence the triangle formed by the three given lines is an isosceles triangle.Option 2 : \(\sqrt 3 {\rm{\;x}} + {\rm{y}} = 10\)

__Concept:__

Equation of a Straight Line in Normal Form: **x cos α + y sin α = p**,

Where p be the length of the normal drawn from the origin to a line, which subtends an angle α with the positive direction of x-axis as follows.

__Calculation:__

Here p = 5

And α = 30°

Now, equation of line,

x cos α + y sin α = p

⇒ x cos 30° + y sin 30° = 5

⇒ x (√3/2) + y (1/2) = 5

⇒ √3x + y = 10

Option 3 : (1, 1)

__Concept:__

All the sides of an equilateral triangle are equal.

__Calculation:__

Let A = (-1, -1), B = ~~\(\left( -\sqrt{3},~\sqrt{3} \right)\)~~ and C = (x, y)

∵ they are vertices of an equilateral triangle ⇒ AB = BC = AC.

\(A{{B}^{2}}={{\left( -\sqrt{3}+1 \right)}^{2}}+{{\left( \sqrt{3}+1 \right)}^{2}}=8~units\)

Similarly, \(B{{C}^{2}}={{\left( x+\sqrt{3} \right)}^{2}}+{{\left( y-\sqrt{3} \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2\sqrt{3}~\left( x-y \right)+6=8\) ----(1)

\(A{{C}^{2}}={{\left( x+1 \right)}^{2}}+{{\left( y+1 \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2\left( x+y \right)+2=8\) ----(2)

From (1) and (2), we get

\(\Rightarrow ~2\sqrt{3}~\left( x-y \right)+6=~2\left( x+y \right)+2\)

⇒ x = y = 1.Equation of the line perpendicular to x - 2y = 1 and passing through (1, 1) is:

Option 4 : y = -2x + 3

__Concept:__

- If two lines y = m
_{1}x + c_{1}and y = m2x + c2 are perpendicular to each other, then m_{1}×m_{2}= -1. - If a point P(a, b) lies on a curve f(x, y) = 0, then f(a, b) = 0.

**Calculation:**

Let us first find out the slope (m_{2}) of the second line.

The equation of the first line is x - 2y = 1.

It can be written as \(\rm y=\frac{1}{2}x-\frac{1}{2}\). Therefore, \(\rm m_1=\frac{1}{2}\).

Now, \(\rm m_1\times m_2=-1\)

⇒ \(\rm \frac{1}{2}\times m_2=-1\)

⇒ m_{2} = -2

∴ The equation of the second line can be written as y = -2x + c.

Since this line passes through (1, 1), we must have:

1 = (-2)(1) + c

⇒ c = 3.

Hence, the equation of the line is **y = -2x + 3**.

__Additional Information__

If two lines y = m1x + c1 and y = m2x + c2 are parallel to each other, then m1 = m2.

In the other form, for two lines a_{1}x + b_{1}y = c_{1} and a2x + b2y = c2:

- If the lines are
**parallel**, then \(\rm \frac{a_1}{b_1}=\frac{a_2}{b_2}\Rightarrow a_1b_2-a_2b_1=0\). - If the lines are
**perpendicular**, then \(\rm \frac{a_1}{b_1}=-\frac{b_2}{a_2}\Rightarrow a_1a_2+b_1b_2=0\).

Option 1 : \(3 \pm 2\sqrt {15}\)

__ CONCEPT__:

Let A (x_{1}, y_{1}) and B (x_{2}, y_{2}) be any two points in the XY – plane, then the distance between A and B is given by:\(\left| {AB} \right| = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \)

__ CALCULATION__:

Given: The distance between the points (3, 4) and (a, 2) is 8 units

Here, we have to find the value of a.

As we know that, the distance between two points A (x1, y1) and B (x2, y2) is given by \(\left| {AB} \right| = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \)

⇒ \(\sqrt {{{\left( {{a} - {3}} \right)}^2} + {{\left( {{2} - {4}} \right)}^2}} = 8\)

By squaring both the sides we get

⇒ (a - 3)^{2} + 4 = 64

⇒ a^{2} + 9 - 6a - 60 = 0

⇒ a^{2} - 6a - 51 = 0

⇒ \(a = \frac{{6 \pm \sqrt {240} }}{2} = 3 \pm 2\sqrt {15} \)

Hence, **option A** is the correct answer.

Option 1 : 30°

__Concept:__

**Angle between two lines:** The angle θ between the lines having slope m_{1} and m_{2} is given by

\(\tan {\rm{\theta }} = {\rm{\;}}\left| {\frac{{{{\rm{m}}_2} - {{\rm{m}}_1}}}{{1 + {\rm{\;}}{{\rm{m}}_1}{\rm{\;}}{{\rm{m}}_2}}}} \right|\)

__Calculation:__

Given: y - √3x – 5 = 0 & √3y – x + 6 = 0

y - √3x – 5 = 0

⇒ y = √3x + 5

So, slope of line, m_{1} = √3

√3y – x + 6 = 0

\(\Rightarrow {\rm{y}} = \frac{{\rm{x}}}{{\sqrt 3 }} - \frac{6}{{\sqrt 3 }}\)

So, slope of the line, m_{2} = \(\frac{1}{{\sqrt 3 }}\)

Let θ be the acute angle between the lines.

\(\tan {\rm{\theta }} = \left| {\frac{{{{\rm{m}}_1} - {{\rm{m}}_2}}}{{1 + {{\rm{m}}_1}{{\rm{m}}_2}}}} \right|\)

\(\Rightarrow \tan {\rm{\theta }} = \left| {\frac{{\sqrt 3 - \frac{1}{{\sqrt 3 }}}}{{1 + \sqrt 3 \times \frac{1}{{\sqrt 3 }}}}} \right|\)

\( \Rightarrow \tan {\rm{\theta }} = \left| {\frac{{\frac{2}{{\sqrt 3 }}}}{2}} \right|\)

\(\Rightarrow \tan {\rm{\theta }} = \frac{1}{{\sqrt 3 }}\)

⇒ θ = 30°

Option 2 : 3x - 2y - 6 = 0

__ CONCEPT__:

If a line cuts the intercepts a and b on the X and Y – axis respectively, then the equation of such line is given as: \(\frac{x}{a} + \frac{y}{b} = 1\)

__ CALCULATION__:

Given: The line makes intercepts of length 2 and - 3 on the line x-axis and the y-axis respectively.

As we know that intercept form of a line is given by: \(\frac{x}{a} + \frac{y}{b} = 1\)

Here, a = 2 and b = - 3

So, the required equation of line is: \(\frac{x}{2} - \frac{y}{3} = 1\)

⇒ 3x - 2y - 6 = 0

Hence, **option B** is the correct answer.

Option 2 : 2x + y = 7

__Concept:__

**Line:** Line in slope form is given as,

\(\frac{{\rm{x}}}{{\rm{a}}} + \frac{{\rm{y}}}{{\rm{b}}} = 1\)

where a is x-intercept and b is y-intercept.

__Calculation:__

Let the equation of line be \(\frac{{\rm{x}}}{{\rm{a}}} + \frac{{\rm{y}}}{{\rm{b}}} = 1\) ----(1)

Where a is x-intercept and b is y-intercept.

The line is passing through (2,3). So, (2,3) should satisfy the equation of line.

Put x = 2 & y = 3 in equation (1)

\(\frac{2}{{\rm{a}}} + \frac{3}{{\rm{b}}} = 1\) ----(2)

Now, we have given that the intercept on the positive y-axis equal to twice its intercept on the positive x-axis.

So, b = 2a

Put b = 2a in equation 2

\(\frac{2}{{\rm{a}}} + \frac{3}{{2{\rm{a}}}} = 1\)

⇒ 2 + 3/2 = a

⇒ a = 7/2

Now, b = 2a = 7

So, equation of line will be,

\(\frac{{\rm{x}}}{{\frac{7}{2}}} + \frac{{\rm{y}}}{7} = 1\)

⇒ 2x + y = 7

Which of the following represents a reflex angle?

Option 4 : (iii) only

An acute angle measures between 0° and 90°, whereas a right angle is exactly equal to 90°.

An angle greater than 90° but less than 180° is called an obtuse angle. Also, recall that a straight angle is equal to 180°.

An angle that is greater than 180° but less than 360° is called a reflex angle.

Further, two angles whose sum is 90° are called complementary angles, and two angles whose sum is 180° are called supplementary angles.

Option 1 : PQ

∠P + ∠Q + ∠R = 180° ( Sum of angles of a triangle)

∠P = 60°, ∠Q = 50°

⇒ ∠R = 70°

In any triangle, the side opposite to the largest (greatest) angle is longest.

The side opposite to ∠R i.e. PQ will be longest.

Option 3 : x + y = 3

**Concept:**

The** slope of a line** passing through the distinct points (x_{1}, y_{1}) and (x_{2}, y_{2}) is \(\rm m = \frac{y_2- y_1}{x_2 - x_1}\)

**Equation of line** is (y – y_{1}) = m(x – x_{1})

**Calculation:**

Given:

Straight-line passing through (5, -2) and (-4, 7)

slope of a line = m = \(\frac{7- (-2)}{-4 - 5} = \frac{9}{-9}=-1\)

Now, the equation of a line is,

y - (-2) = -1 × (x - 5)

⇒ y + 2 = -x + 5

⇒ x + y = 3

Consider the following statements:

1) The length p of the perpendicular form the origin to the line ax + by = c satisfies the origin to the relation \({{\rm{p}}^2} = \frac{{{{\rm{c}}^2}}}{{{{\rm{a}}^2} + {{\rm{b}}^2}}}\)

2) The length p the perpendicular from the origin to the line \(\frac{{\rm{x}}}{{\rm{a}}} + \frac{{\rm{y}}}{{\rm{b}}} = 1\) satisfies the relation \(\frac{1}{{{{\rm{p}}^2}}} = \frac{1}{{{{\rm{a}}^2}}} + \frac{1}{{{{\rm{b}}^2}}}\)

3) The length p of the perpendicular from the origin to the line y = mx + c satisfies the relation \(\frac{1}{{{{\rm{p}}^2}}} = \frac{{1 + {{\rm{m}}^2} + {{\rm{c}}^2}}}{{{{\rm{c}}^2}}}\)

Which of the above is/are correct?Option 3 : 1 and 2 only

__Concept:__

**Perpendicular distance **from a point on a line**:**

Let a point be (x_{1}, y_{1}) and equation of line be ax + by + c = 0 then:

Perpendicular distance \(= \frac{{\left| {{\rm{a}}{{\rm{x}}_1}{\rm{\;}} + {\rm{\;b}}{{\rm{y}}_1}{\rm{\;}} + {\rm{\;c}}} \right|}}{{\sqrt {{{\rm{a}}^2}{\rm{\;}} + {\rm{\;}}{{\rm{b}}^2}} }}\)

__Calculation:__

For **statement (1),**

Point is (0,0) and equation of line is ax + by - c = 0.

So, \({\rm{p}} = \frac{{\left| { - {\rm{c}}} \right|}}{{\sqrt {{{\rm{a}}^{2{\rm{\;}}}} + {\rm{\;}}{{\rm{b}}^2}} }}\)

\(\Rightarrow {\rm{\;p}} = \frac{{\rm{c}}}{{\sqrt {{{\rm{a}}^{2{\rm{\;}}}} + {\rm{\;}}{{\rm{b}}^2}} }}\)

Squaring both sides

\(\Rightarrow {{\rm{p}}^2} = \frac{{{{\rm{c}}^2}}}{{{{\rm{a}}^2} + {{\rm{b}}^2}}}\)

So, **first statement **is correct.

For **statement (2),**

Point is (0,0) and equation of line is \(\frac{{\rm{x}}}{{\rm{a}}} + \frac{{\rm{y}}}{{\rm{b}}} = 1\)

Equation of line will be bx + ay = ab

⇒ bx + ay - ab = 0

So, \({\rm{p}} = \frac{{\left| { - {\rm{ab}}} \right|}}{{\sqrt {{{\rm{b}}^2}{\rm{\;}} + {\rm{\;}}{{\rm{a}}^2}} }}\)

\(\Rightarrow {\rm{p}} = \frac{{{\rm{ab}}}}{{\sqrt {{{\rm{b}}^2}{\rm{\;}} + {\rm{\;}}{{\rm{a}}^2}} }}\)

Squaring both sides

\(\Rightarrow {{\rm{p}}^2} = \frac{{{{\rm{a}}^2}{{\rm{b}}^2}}}{{{{\rm{a}}^2} + {{\rm{b}}^2}}}\)

Taking reciprocal of both sides,

\(\Rightarrow \frac{1}{{{{\rm{p}}^2}}} = \frac{{{{\rm{a}}^2} + {{\rm{b}}^2}}}{{{{\rm{a}}^2}{{\rm{b}}^2}}}\)

\(\Rightarrow \frac{1}{{{{\rm{p}}^2}}} = \frac{1}{{{{\rm{b}}^2}}} + \frac{1}{{{{\rm{a}}^2}}}\)

So, **second statement **is correct.

For **statement (3),**

Point is (0,0) and equation of line is y - mx - c = 0

So, \({\rm{p}} = \frac{{\left| { - {\rm{c}}} \right|}}{{\sqrt {{{\rm{m}}^2} + {1^2}} }}\)

Squaring both sides,

\(\Rightarrow {{\rm{p}}^2} = \frac{{{{\rm{c}}^2}}}{{{{\rm{m}}^2} + 1}}\)

Taking reciprocal of both sides,

\(\Rightarrow \frac{1}{{{{\rm{p}}^2}}} = \frac{{{{\rm{m}}^2} + 1}}{{{{\rm{c}}^2}}}\)

So,Option 4 : a = -1

**Concept:**

The lines a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b2y + c2 = 0 are perpendicular then product of slope is -1.

⇒ a_{1} a_{2} + b_{1} b_{2 }= 0

**Calculations:**

Given, the lines x + (a - 1)y + 1 = 0 and 2x + a2y - 1 = 0 are perpendicular.

⇒ \(\rm 1\times 2\; + (a - 1)\times a^2 = 0\)

⇒ \(\rm a^3 - a^2 + 2 = 0\)

⇒ \(\rm a^3 + a^2- 2a^2 - 2a + 2a + 2 = 0\)

⇒ \(\rm a^2(a+ 1)-2a (a+1)+ 2(a+1) = 0\)

⇒ \(\rm (a+1)(a^2-2a + 2) = 0\)

⇒ \(\rm (a+1)= 0\)

⇒ \(\rm a= -1\)

Option 4 : Either x + y =5 or 9x + 4y = 30

__Concept:__

- The
**intercept form of the line**⇔ \(\frac{x}{a} + \frac{y}{b} = 1\)

Where a is the x- intercept and b is the y- intercept

**Section Formula:**Section formula is used to determine the coordinate of a point that divides a line into two parts such that ratio of their length is m : n- Let P and Q be the given two points (x
_{1}, y_{1}) and (x_{2}, y_{2}) respectively and M(x, y) be the point dividing the line- segment PQ internally in the ratio m: n

**Internal Section Formula:**When the line segment is divided internally in the ration m: n, we use this formula. ⇔ \(\left( {x,{\rm{\;}}y} \right) = \left( {\frac{{m{x_2}{\rm{\;}} + {\rm{\;}}n{x_1}}}{{m\; + {\rm{\;}}n}},\frac{{m{y_2}{\rm{\;}} + {\rm{\;}}n{y_1}}}{{m\; + {\rm{\;}}n}}} \right)\)**External Section Formula:**When the point M lies on the external part of the line segment. ⇔ \(\left( {x,{\rm{\;}}y} \right) = \left( {\frac{{m{x_2} - {\rm{\;}}n{x_1}}}{{m\; - {\rm{\;}}n}},\frac{{m{y_2} - {\rm{\;}}n{y_1}}}{{m\; - {\rm{\;}}n}}} \right)\)

__Calculation:__

Intercept form of the line is \(\frac{x}{a} + \frac{y}{b} = 1\)

At x axis point is A (a, 0)

At y axis point is B (0, b)

**Case 1:** Given point P (2, 3) divides A and B in ratio of m : n = 3 : 2

Applying Internal Section Formula,

We know that \(\left( {x,{\rm{\;}}y} \right) = \left( {\frac{{m{x_2}{\rm{\;}} + {\rm{\;}}n{x_1}}}{{m\; + {\rm{\;}}n}},\frac{{m{y_2}{\rm{\;}} + {\rm{\;}}n{y_1}}}{{m\; + {\rm{\;}}n}}} \right)\)

\(\Rightarrow \;\left( {2,\;3} \right) = \;\left( {\frac{{3\left( 0 \right) + \;2\left( a \right)}}{{3 + 2}},\frac{{3\left( b \right) + 2\left( 0 \right)}}{{3 + 2}}} \right)\)

\( \Rightarrow \;\left( {2,\;3} \right) = \;\left( {\frac{{2a}}{5},\frac{{3b}}{5}} \right)\)

Now,

\(\frac{{2a}}{5} = 2\;and\;\frac{{3b}}{5} = 3\)

⇒ a = 5 and b = 5

∴ Equation of line is \(\frac{x}{5} + \frac{y}{5} = 1\) ⇒ x + y = 5

**Case 2:** Given point P (2, 3) divides A and B in ratio of m : n = 2 : 3

\( \Rightarrow \;\left( {2,\;3} \right) = \;\left( {\frac{{2\left( 0 \right) + \;3\left( a \right)}}{{3 + 2}},\frac{{2\left( b \right) + 3\left( 0 \right)}}{{3 + 2}}} \right)\)

\( \Rightarrow \;\left( {2,\;3} \right) = \;\left( {\frac{{3a}}{5},\frac{{2b}}{5}} \right)\)

Now,

\(\frac{{3a}}{5} = 2\;and\;\frac{{2b}}{5} = 3\)

⇒ a = 10/3 and b = 15/2

∴ Equation of line is \(\frac{x}{\left( {}^{10}\!\!\diagup\!\!{}_{3}\; \right)}+\frac{y}{\left( {}^{15}\!\!\diagup\!\!{}_{2}\; \right)}=1\)

\(\Rightarrow {\rm{\;}}\frac{{3x}}{{10}} + \frac{2}{{15}} = 1\)

⇒ 9x + 4y = 30

__Shortcut Method:__

- Check through the options: Substitute the point (2, 3) in all options

Option 1 : \(\frac{{12\sqrt {17} }}{5}\)

__Concept:__

**Section formula: **Let P(x, y) divides AB in ratio (m : n), where A ≡ (x_{1}, y_{1}) and B ≡ (x_{2}, y_{2}) then:

\({\rm{x}} = \frac{{{\rm{m}}{{\rm{x}}_2}{\rm{\;}} \pm {\rm{\;n}}{{\rm{x}}_1}}}{{{\rm{m\;}} \pm {\rm{\;n}}}}\) & \({{\rm{y}}_1} = \frac{{{\rm{m}}{{\rm{y}}_2}{\rm{\;}} \pm {\rm{\;n}}{{\rm{y}}_1}}}{{{\rm{m\;}} \pm {\rm{\;n}}}}\)

+ve for internal division & -ve for external division.

**Distance formula: **Let A ≡ (x_{1}, y_{1}) and B ≡ (x_{2}, y_{2}), distance between A and B is given as:

\({\rm{AB}} = \sqrt {{{\left( {{{\rm{x}}_2} - {{\rm{x}}_2}} \right)}^2} + {{\left( {{{\rm{x}}_2} - {{\rm{x}}_2}} \right)}^2}} \)

__Calculation:__

Let A ≡ (4, 3) and B ≡ (5, 7)

Let P(x_{1}, y_{1}) divides AB externally and Q(x_{2}, y_{2}) divides AB internally.

Given: ratio of division 2 : 3

To find: length of PQ

External division:

Using **section formula,**

\({{\rm{x}}_1} = \frac{{2 \times 5{\rm{\;}} - {\rm{\;}}3 \times 4}}{{2{\rm{\;}} - {\rm{\;}}3}}\) & \({{\rm{y}}_1} = \frac{{2 \times 7{\rm{\;}} - {\rm{\;}}3 \times 3}}{{2{\rm{\;}} - {\rm{\;}}3}}\)

⇒ x_{1} = 2 & y_{1} = -5

Internal division;

Using **section formula,**

\({{\rm{x}}_2} = \frac{{2 \times 5{\rm{\;}} + {\rm{\;}}3 \times 4}}{{2{\rm{\;}} + {\rm{\;}}3}}\) & \({{\rm{y}}_2} = \frac{{2 \times 7{\rm{\;}} + {\rm{\;}}3 \times 3}}{{2{\rm{\;}} + {\rm{\;}}3}}\)

⇒ x_{2} = 22/5 & y_{2} = 23/5

So, P ≡ (2, -5) & Q ≡ (22/5, 23/5)

\({\rm{PQ}} = \sqrt {{{\left( {{{\rm{x}}_2} - {{\rm{x}}_2}} \right)}^2} + {{\left( {{{\rm{x}}_2} - {{\rm{x}}_2}} \right)}^2}} \)

\(= \sqrt {{{\left( {\frac{{22}}{5} - 2} \right)}^2} + {{\left( {\frac{{23}}{5} - \left( { - 5} \right)} \right)}^2}} \)

\(= \sqrt {{{\left( {\frac{{12}}{5}} \right)}^2} + {{\left( {\frac{{48}}{5}} \right)}^2}} \)

\(= \frac{{12}}{5}\sqrt {1 + {4^2}} \)

\(= \frac{{12\sqrt {17} }}{5}\)