JEE Main 2018 (15 April Evening)

LEVELJEE Main

Animated Solution for Mathematics - Straight Lines: The sides of a rhombus ABCD are parallel to the lines, $x - y + 2 = 0$ and $7 x - y + 3 = 0$ . If the diagonals of the rhombus intersect at P(1, 2) and the vertex A (different from the origin) is on the y-axis, then the ordinate of A is :-

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the rhombus. The diagonals are not just lines; they are the angle bisectors of the sides. This is the geometric heart of the problem. If you can find the directions of these bisectors, you have found the directions of the diagonals.

Hint 2

Hint 2 (The Arsenal): To find the angle bisectors of two lines with slopes

m_{1}

and

m_{2}

, you can use the direction vectors of the lines. Normalize these vectors to unit vectors

\overset{u}{^}_{1}

and

\overset{u}{^}_{2}

. The angle bisectors will then be in the directions of

\overset{u}{^}_{1} + \overset{u}{^}_{2}

and

\overset{u}{^}_{1} - \overset{u}{^}_{2}

Hint 3

Hint 3 (The Execution): Once you have the slopes of the diagonals, use the point-slope form with the intersection point

P (1, 2)

to find the equations of the two diagonals. Then, substitute

x = 0

into these equations to find the possible

y

-coordinates of vertex

A

, remembering to reject the origin

(0, 0)

as per the problem statement.

Visualized Solution

The Rhombus and its Diagonals

Let's visualize the rhombus $A B C D$ .
The diagonals of a rhombus intersect at $P (1, 2)$ .
The sides of the rhombus are parallel to the given lines.

Slopes of the Sides

The sides are parallel to two given lines:
$L_{1} : x - y + 2 = 0 ⟹ m_{1} = 1$
$L_{2} : 7 x - y + 3 = 0 ⟹ m_{2} = 7$

Property of Rhombus Diagonals

In a rhombus, the diagonals bisect the interior angles.
Therefore, the diagonals are parallel to the angle bisectors of the sides.

Finding Diagonal Directions

Let's find the direction vectors of the sides.
For $m_{1} = 1$ , direction vector $v_{1} = (1, 1)$ .
For $m_{2} = 7$ , direction vector $v_{2} = (1, 7)$ .

Unit Vectors of Sides

Convert direction vectors to unit vectors:
$\overset{u}{^}_{1} = \frac{1}{1 ^{2} + 1 ^{2}} (1, 1) = \frac{1}{2} (1, 1)$
$\overset{u}{^}_{2} = \frac{1}{1 ^{2} + 7 ^{2}} (1, 7) = \frac{1}{5 2} (1, 7)$

Slopes of the Diagonals

The angle bisectors are along $\overset{u}{^}_{1} \pm \overset{u}{^}_{2}$ .
$\overset{u}{^}_{1} + \overset{u}{^}_{2} = \frac{1}{5 2} (5, 5) + \frac{1}{5 2} (1, 7) = \frac{1}{5 2} (6, 12) \propto (1, 2)$
$\overset{u}{^}_{1} - \overset{u}{^}_{2} = \frac{1}{5 2} (5, 5) - \frac{1}{5 2} (1, 7) = \frac{1}{5 2} (4, - 2) \propto (2, - 1)$
Slopes of diagonals: $M_{1} = 2$ and $M_{2} = - \frac{1}{2}$ .

Equations of the Diagonals

We know the diagonals pass through $P (1, 2)$ .
Equation 1: $y - 2 = 2 (x - 1)$
Equation 2: $y - 2 = - \frac{1}{2} (x - 1)$

Simplifying Diagonal Equations

Simplifying Equation 1: $y - 2 = 2 x - 2 ⟹ y = 2 x$
Simplifying Equation 2: $2 y - 4 = - x + 1 ⟹ x + 2 y = 5$

Locating Vertex $A$

The problem states vertex $A$ lies on the $y$ -axis.
Therefore, the $x$ -coordinate of $A$ is $0$ , so $A = (0, y_{A})$ .
Vertex $A$ must lie on one of the diagonals.

Testing the First Diagonal

Let's test if $A$ lies on the first diagonal: $y = 2 x$ .
Substitute $x = 0$ into $y = 2 x$ .

Evaluating the First Case

$y = 2 (0) ⟹ y = 0$ .
This gives the point $(0, 0)$ , which is the origin.
But the problem states $A$ is different from the origin.
So, $A$ cannot lie on $y = 2 x$ .

Testing the Second Diagonal

Since $A$ is not on $y = 2 x$ , it must lie on the second diagonal: $x + 2 y = 5$ .
Substitute $x = 0$ into $x + 2 y = 5$ .

Calculating the Ordinate of $A$

$0 + 2 y = 5$
$2 y = 5$
$y = \frac{5}{2}$
The coordinates of $A$ are $(0, \frac{5}{2})$ .

Final Conclusion

The ordinate of vertex $A$ is $\frac{5}{2}$ .
Key Concept: Diagonals of a rhombus bisect the interior angles.
Pro Tip: Always check the constraints given in the problem (like $A$ is not the origin).

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The Sigma Insight: Angle Between Two Lines

Solution Diagram

Analyzing the Setup

Welcome, fellow traveler in the world of coordinate geometry! Today, we are going to dissect a beautiful problem involving a rhombus.

Imagine you are standing on a coordinate plane, and before you lies a rhombus

A B C D

. It is defined by its sides being parallel to two specific lines:

L_{1} : x - y + 2 = 0

L_{2} : 7 x - y + 3 = 0

We are told the diagonals intersect at

P (1, 2)

, and vertex

A

resides on the

y

-axis. Our mission is to find the ordinate of

A

The Heart of the Rhombus

The intersection point

P (1, 2)

is the heart of our shape. In any rhombus, the diagonals are the angle bisectors of the interior angles.

This is a powerful property! It means the diagonals are parallel to the angle bisectors of the lines forming the sides. Let's find the slopes of our sides first.

For

L_{1} : x - y + 2 = 0

, the slope

m_{1}

1

. For

L_{2} : 7 x - y + 3 = 0

, the slope

m_{2}

7

The Vector Approach

To find the angle bisectors, we need the direction vectors of these lines. For a line with slope

m

, the direction vector is

(1, m)

So, for our sides, we have

v_{1} = (1, 1)

and

v_{2} = (1, 7)

. Now, let's normalize these into unit vectors:

\overset{u}{^}_{1} = \frac{1}{2} (1, 1)

\overset{u}{^}_{2} = \frac{1}{5 2} (1, 7)

The directions of the angle bisectors are given by the sum and difference of these unit vectors:

\overset{u}{^}_{1} + \overset{u}{^}_{2}

and

\overset{u}{^}_{1} - \overset{u}{^}_{2}

When we add them, we get a vector proportional to

(6, 12)

, which simplifies to

(1, 2)

. This gives us a slope of

M_{1} = 2

When we subtract them, we get a vector proportional to

(4, - 2)

, which simplifies to

(2, - 1)

, giving us a slope of

M_{2} = - \frac{1}{2}

Notice the elegance here:

M_{1} \times M_{2} = 2 \times (- \frac{1}{2}) = - 1

. The diagonals are perpendicular, just as they should be in a rhombus!

The Final Reveal

Now that we have the slopes of the diagonals, we use the point-slope form, knowing they pass through

P (1, 2)

The first diagonal is

y - 2 = 2 (x - 1)

, which simplifies to

y = 2 x

. The second diagonal is

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

, which simplifies to

x + 2 y = 5

Since vertex

A

lies on the

y

-axis, its

x

-coordinate is

0

. If

A

were on the first diagonal,

y = 2 (0) = 0

, giving us the origin

(0, 0)

But the problem implies

A

is a distinct vertex. Thus,

A

must lie on the second diagonal.

Substituting

x = 0

into

x + 2 y = 5

, we get

2 y = 5

, or

y = \frac{5}{2}

The ordinate of

A

is $\frac{5}{2}$ .

Common Traps & FAQs

Wait, why do we use unit vectors to find the angle bisectors? Can't we just use the formula for the angle bisector of two lines?

You absolutely can use the standard formula for the angle bisector of two lines,

L_{1}

and

L_{2}

, which is

\frac{L _{1}}{A _{1}^{2} + B _{1}^{2}} = \pm \frac{L _{2}}{A _{2}^{2} + B _{2}^{2}}

. However, using unit vectors is often more intuitive and less prone to algebraic errors in coordinate geometry. It directly gives you the direction vector of the bisector, which is the slope we need!

Why do we reject the origin

(0, 0)

? The math gave us

y = 0

as a solution.

That is a great catch! In competitive exams like JEE, always re-read the problem statement carefully. The question explicitly states that vertex

A

is 'different from the origin'. This is a constraint designed to test your attention to detail. Since

A

cannot be

(0, 0)

, we must discard that result.

Why can't we just use the slopes of the sides to find the vertices directly?

The slopes of the sides only tell us the orientation of the rhombus, not the position of the vertices. The diagonals are the key because they pass through the intersection point

P (1, 2)

and are perpendicular to each other. Without the diagonal equations, we have no way to constrain the coordinates of the vertices.

Animated Solution for Mathematics - Straight Lines: The sides of a rhombus ABCD are parallel to the lines, $x - y + 2 = 0$ and $7 x - y + 3 = 0$ . If the diagonals of the rhombus intersect at P(1, 2) and the vertex A (different from the origin) is on the y-axis, then the ordinate of A is :-

Select Answer:

Visualized Solution

The Rhombus and its Diagonals

Slopes of the Sides

Property of Rhombus Diagonals

Finding Diagonal Directions

Unit Vectors of Sides

Slopes of the Diagonals

Equations of the Diagonals

Simplifying Diagonal Equations

Locating Vertex $A$

Testing the First Diagonal

Evaluating the First Case

Testing the Second Diagonal

Calculating the Ordinate of $A$

Final Conclusion

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

The Heart of the Rhombus

The Vector Approach

The Final Reveal

Similar Questions

Two sides of a rhombus $A B C D$ are parallel to the lines $y = x + 2$ and $y = 7 x + 3$ . If the diagonals of the rhombus intersect at the point $(1, 2)$ and the vertex $A$ is on the y-axis, find possible co-ordinates of $A$ .

Let $A (1, 0), B (2, - 1)$ and $C (\frac{7}{3}, \frac{4}{3})$ be three points. If the equation of the bisector of the angle $A B C$ is $α x + β y = 5$ , then the value of $α^{2} + β^{2}$ is

The diagonals of a parallelogram $P QR S$ are along the lines $x + 3 y = 4$ and $6 x - 2 y = 7$ . Then $P QR S$ must be a

The distance between the two points $A$ and $A^{'}$ which lie on $y = 2$ such that both the line segments $A B$ and $A^{'} B$ (where $B$ is the point $(2, 3)$ ) subtend angle $\frac{π}{4}$ at the origin, is equal to :

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 one of the lines of $m y^{2} + (1 - m^{2}) x y - m x^{2} = 0$ is a bisector of the angle between the lines $x y = 0$ , then $m$ is

If the two lines $x + (a - 1) y = 1$ and $2 x + a^{2} y = 1 (a \in R - {0, 1})$ are perpendicular, then the distance of their point of intersection from the origin is :-

Consider a triangle $A B C$ having the vertices $A (1, 2)$ , $B (α, β)$ and $C (γ, δ)$ and angles $∠ A B C = \frac{π}{6}$ and $∠ B A C = \frac{2 π}{3}$ . If the points $B$ and $C$ lie on the line $y = x + 4$ , then $α^{2} + γ^{2}$ is equal to ________

Select Answer:

Visualized Solution

The Rhombus and its Diagonals

Slopes of the Sides

Property of Rhombus Diagonals

Finding Diagonal Directions

Unit Vectors of Sides

Slopes of the Diagonals

Equations of the Diagonals

Simplifying Diagonal Equations

Locating Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Testing the First Diagonal

Evaluating the First Case

Testing the Second Diagonal

Calculating the Ordinate of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Final Conclusion

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

The Heart of the Rhombus

The Vector Approach

The Final Reveal

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the two lines x+(a−1)y=1 and 2x+a2y=1(a∈R−{0,1}) are perpendicular, then the distance of their point of intersection from the origin is :-

Locating Vertex $A$

Calculating the Ordinate of $A$

Let $A (1, 0), B (2, - 1)$ and $C (\frac{7}{3}, \frac{4}{3})$ be three points. If the equation of the bisector of the angle $A B C$ is $α x + β y = 5$ , then the value of $α^{2} + β^{2}$ is

The diagonals of a parallelogram $P QR S$ are along the lines $x + 3 y = 4$ and $6 x - 2 y = 7$ . Then $P QR S$ must be a

If one of the lines of $m y^{2} + (1 - m^{2}) x y - m x^{2} = 0$ is a bisector of the angle between the lines $x y = 0$ , then $m$ is

If the two lines $x + (a - 1) y = 1$ and $2 x + a^{2} y = 1 (a \in R - {0, 1})$ are perpendicular, then the distance of their point of intersection from the origin is :-