JEE Main 2021 (March)

LEVELJEE Main

Animated Solution for Mathematics - Straight Lines: The equation of one of the straight lines which passes through the point (1,3) and makes an angles $tan^{- 1} (2)$ with the straight line, $y + 1 = 32 x$ is

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Visualized Solution

Visualizing the Setup

Given point: $P (1, 3)$
Given line $L_{1}$ : $y + 1 = 32 x$
Angle between lines: $θ = tan^{- 1} (2)$

Finding the Reference Slope $m_{1}$

Rewrite $L_{1}$ in slope-intercept form: $y = 32 x - 1$
Comparing with $y = m x + c$ , we get slope $m_{1} = 32$

The Angle Formula

Let the slope of the required line be $m$ .
Formula: $tan θ = \frac{m - m _{1}}{1 + m m _{1}}$

Setting up the Equation

Given $tan θ = 2$ and $m_{1} = 32$
Substitute values: $2 = \frac{m - 3 2}{1 + m ( 3 2 )}$

Squaring Both Sides

Square both sides to remove modulus:
$2 = (\frac{m - 3 2}{1 + 3 2 m})^{2}$
$2 (1 + 32 m)^{2} = (m - 32)^{2}$

Expanding the Terms

Expand left side: $2 (1 + 18 m^{2} + 62 m)$
Expand right side: $m^{2} + 18 - 62 m$

Forming the Quadratic Equation

Equate expansions: $2 + 36 m^{2} + 122 m = m^{2} + 18 - 62 m$
Rearrange terms: $35 m^{2} + 182 m - 16 = 0$

Solving for $m$

Using quadratic formula $m = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ :
$m = \frac{- 18 2 \pm ( 18 2 ) ^{2} - 4 ( 35 ) ( - 16 )}{2 ( 35 )}$

Simplifying the Discriminant

Discriminant $D = 648 + 2240 = 2888$
$2888 = 8 \times 361 = 22 \times 19 = 382$
$m = \frac{- 18 2 \pm 38 2}{70}$

Calculating Slope Values

Case 1 (Positive): $m = \frac{20 2}{70} = \frac{2 2}{7}$
Case 2 (Negative): $m = \frac{- 56 2}{70} = - \frac{4 2}{5}$

Finding the Line Equation

Using $m = - \frac{4 2}{5}$ and point $(1, 3)$ :
Point-slope form: $y - 3 = - \frac{4 2}{5} (x - 1)$
$5 (y - 3) = - 42 (x - 1)$

Final Answer and Summary

$5 y - 15 = - 42 x + 42$
Rearrange: $42 x + 5 y - 15 - 42 = 0$
Final Equation: $42 x + 5 y - (15 + 42) = 0$

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The Sigma Insight: Angle Between Two Lines

Solution Diagram

Analyzing the Setup

You are standing at the point

P (1, 3)

on a coordinate plane. You are tasked with finding the equation of a line passing through

P

that intersects the line

L_{1}

at an angle

θ = tan^{- 1} (2)

The line

L_{1}

is defined by the equation

y + 1 = 32 x

Unmasking the Slope

To determine the slope of

L_{1}

, we rewrite the equation in the slope-intercept form

y = m x + c

. By isolating

y

, we obtain:

y = 32 x - 1

Thus, the slope of the existing line is

m_{1} = 32

. This value serves as the anchor for our geometric calculations.

The Bridge of Tangents

The relationship between the angle

θ

between two lines and their respective slopes

m

and

m_{1}

is given by the formula:

tan θ = \frac{m - m _{1}}{1 + m m _{1}}

Given that

tan θ = 2

, we substitute our known values into the equation:

2 = \frac{m - 3 2}{1 + 3 2 m}

The Algebraic Transformation

To solve for

m

, we square both sides of the equation to eliminate the modulus:

2 = (\frac{m - 3 2}{1 + 3 2 m})^{2}

Expanding the numerator and denominator, we get:

2 = \frac{m ^{2} - 6 2 m + 18}{18 m ^{2} + 6 2 m + 1}

Cross-multiplying yields

2 (18 m^{2} + 62 m + 1) = m^{2} - 62 m + 18

. Simplifying this expression results in the quadratic equation:

35 m^{2} + 182 m - 16 = 0

Final Calculation

We solve for

m

using the quadratic formula

m = \frac{- b \pm b ^{2} - 4 a c}{2 a}

. The discriminant is calculated as:

D = (182)^{2} - 4 (35) (- 16) = 648 + 2240 = 2888 = (382)^{2}

This provides two possible slopes:

m = \frac{- 18 2 \pm 38 2}{70}

The two potential slopes are

m = \frac{2 2}{7}

and

m = - \frac{4 2}{5}

. Using the slope

m = - \frac{4 2}{5}

and the point-slope form

y - y_{1} = m (x - x_{1})

at point

P (1, 3)

, we have:

y - 3 = - \frac{4 2}{5} (x - 1)

Multiplying by

5

and rearranging the terms, we arrive at the final equation:

$42 x + 5 y - (15 + 42) = 0$

Common Traps & FAQs

Why do we use the modulus in the angle formula?

The angle between two lines is defined as the acute angle, but algebraically, the formula

\frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}

can yield a negative value depending on which line you subtract from which. The modulus ensures we are calculating the magnitude of the tangent of the angle, which is always positive for acute angles.

Wait, why are there two possible slopes?

Think about it geometrically! If you are at a point and you draw a line at a specific angle to another line, you can rotate your line clockwise or counter-clockwise by that angle. Both lines will satisfy the condition. That is why the quadratic equation gives you two roots.

Is squaring both sides always safe?

Squaring is a powerful tool to remove the modulus, but be careful. It can sometimes introduce extraneous roots. However, in this specific case, since we are dealing with a linear relationship between slopes, the quadratic equation correctly captures both valid slopes. Always verify your final slope by plugging it back into the original equation if you are unsure.

Animated Solution for Mathematics - Straight Lines: The equation of one of the straight lines which passes through the point (1,3) and makes an angles $tan^{- 1} (2)$ with the straight line, $y + 1 = 32 x$ is

Select Answer:

Visualized Solution

Visualizing the Setup

Finding the Reference Slope $m_{1}$

The Angle Formula

Setting up the Equation

Squaring Both Sides

Expanding the Terms

Forming the Quadratic Equation

Solving for $m$

Simplifying the Discriminant

Calculating Slope Values

Finding the Line Equation

Final Answer and Summary

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

Unmasking the Slope

The Bridge of Tangents

The Algebraic Transformation

Final Calculation

Similar Questions

A straight line $L$ through the point $(3, - 2)$ is inclined at an angle $6 0^{\circ}$ to the line $3 x + y = 1$ . If $L$ also intersects the x-axis, then the equation of $L$ is

Let $P QR$ be a right angled isosceles triangle, right angled at $P (2, 1)$ . If the equation of the line $QR$ is $2 x + y = 3$ , then the equation representing the pair of lines $P Q$ and $P R$ is

The combined equation of the two lines $a x + b y + c = 0$ and $a^{'} x + b^{'} y + c^{'} = 0$ can be written as $(a x + b y + c) (a^{'} x + b^{'} y + c^{'}) = 0$ . The equation of the angle bisectors of the lines represented by the equation $2 x^{2} + x y - 3 y^{2} = 0$ is

Straight lines $3 x + 4 y = 5$ and $4 x - 3 y = 15$ intersect at the point $A$ . Points $B$ and $C$ are chosen on these two lines such that $A B = A C$ . Determine the possible equations of the line $B C$ passing through the point $(1, 2)$ .

If the line $l_{1} : 3 y - 2 x = 3$ is the angular bisector of the lines $l_{2} : x - y + 1 = 0$ and $l_{3} : α x + β y + 17 = 0$ , then $α^{2} + β^{2} - α - β$ is equal to

If the two lines $l_{1} : \frac{x - 2}{3} = \frac{y + 1}{- 2}, z = 2$ and $l_{2} : \frac{x - 1}{1} = \frac{2 y + 3}{α} = \frac{z + 5}{2}$ are perpendicular, then an angle between the lines $l_{2}$ and $l_{3} : \frac{1 - x}{3} = \frac{2 y - 1}{- 4} = \frac{z}{4}$ is:

If the curve $x^{2} + 2 y^{2} = 2$ intersects the line $x + y = 1$ at two points $P$ and $Q$ , then the angle subtended by the line segment $P Q$ at the origin is:

Let the equation of the pair of lines, $y = p x$ and $y = q x$ , can be written as $(y - p x) (y - q x) = 0$ . Then the equation of the pair of the angle bisectors of the lines $x^{2} - 4 x y - 5 y^{2} = 0$ is:

Select Answer:

Visualized Solution

Visualizing the Setup

Finding the Reference Slope .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m1​

The Angle Formula

Setting up the Equation

Squaring Both Sides

Expanding the Terms

Forming the Quadratic Equation

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Simplifying the Discriminant

Calculating Slope Values

Finding the Line Equation

Final Answer and Summary

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

Unmasking the Slope

The Bridge of Tangents

The Algebraic Transformation

Final Calculation

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let PQR be a right angled isosceles triangle, right angled at P(2,1). If the equation of the line QR is 2x+y=3, then the equation representing the pair of lines PQ and PR is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the line l1​:3y−2x=3 is the angular bisector of the lines l2​:x−y+1=0 and l3​:αx+βy+17=0, then α2+β2−α−β is equal to

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Finding the Reference Slope $m_{1}$

Solving for $m$

A straight line $L$ through the point $(3, - 2)$ is inclined at an angle $6 0^{\circ}$ to the line $3 x + y = 1$ . If $L$ also intersects the x-axis, then the equation of $L$ is

Let $P QR$ be a right angled isosceles triangle, right angled at $P (2, 1)$ . If the equation of the line $QR$ is $2 x + y = 3$ , then the equation representing the pair of lines $P Q$ and $P R$ is

If the line $l_{1} : 3 y - 2 x = 3$ is the angular bisector of the lines $l_{2} : x - y + 1 = 0$ and $l_{3} : α x + β y + 17 = 0$ , then $α^{2} + β^{2} - α - β$ is equal to

If the curve $x^{2} + 2 y^{2} = 2$ intersects the line $x + y = 1$ at two points $P$ and $Q$ , then the angle subtended by the line segment $P Q$ at the origin is: