JEE Main 2020 (7 January Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Straight Lines: Let $A (1, 0), B (6, 2), C (\frac{3}{2}, 6)$ be the vertices of a triangle $A B C$ . If $P$ is a point inside the triangle $A B C$ such that the triangles $A P C, A P B$ and $B P C$ have equal areas, then the length of the line segment $P Q$ , where $Q$ is the point $(- \frac{7}{6}, - \frac{1}{3})$ , is

Q: Is there a trick to the distance calculation? The fractions look intimidating.

I understand the hesitation! When you see fractions like $\frac{17}{6}$ and $-\frac{7}{6}$, it might look messy. But notice that when you subtract them, the denominators are the same, which makes the arithmetic much cleaner. Always look for that simplification before you start squaring large numbers. It is often a sign that you are on the right track!

Enter Numerical Value:

Visualized Solution

Visualizing Triangle $A B C$

Vertices: $A (1, 0)$ , $B (6, 2)$ , $C (\frac{3}{2}, 6)$

Identifying Point $P$

Given: $A r e a (△ A P C) = A r e a (△ A P B) = A r e a (△ B P C)$
This property implies that point $P$ is the Centroid of $△ A B C$ .

Applying the Centroid Formula

Centroid formula: $P (x, y) = (\frac{x _{1} + x _{2} + x _{3}}{3}, \frac{y _{1} + y _{2} + y _{3}}{3})$
Substituting $x$ -coordinates: $x_{P} = \frac{1 + 6 + \frac{3}{2}}{3}$
Substituting $y$ -coordinates: $y_{P} = \frac{0 + 2 + 6}{3}$

Calculating $x$ -coordinate of $P$

$x_{P} = \frac{7 + \frac{3}{2}}{3}$
$x_{P} = \frac{\frac{14 + 3}{2}}{3}$
$x_{P} = \frac{17}{6}$

Calculating $y$ -coordinate of $P$

$y_{P} = \frac{8}{3}$
Point $P \equiv (\frac{17}{6}, \frac{8}{3})$

Defining Point $Q$ and Distance Formula

Point $Q = (- \frac{7}{6}, - \frac{1}{3})$
Distance $P Q = (x_{P} - x_{Q})^{2} + (y_{P} - y_{Q})^{2}$

Substituting Values in Distance Formula

$P Q = (\frac{17}{6} - (- \frac{7}{6}))^{2} + (\frac{8}{3} - (- \frac{1}{3}))^{2}$
$P Q = (\frac{17}{6} + \frac{7}{6})^{2} + (\frac{8}{3} + \frac{1}{3})^{2}$

Simplifying the Coordinate Differences

$Δ x = \frac{17}{6} + \frac{7}{6} = \frac{24}{6} = 4$
$Δ y = \frac{8}{3} + \frac{1}{3} = \frac{9}{3} = 3$
$P Q = 4^{2} + 3^{2}$

Final Calculation

$P Q = 16 + 9$
$P Q = 25$
$P Q = 5$

Summary and Key Takeaway

Key Takeaway: The point dividing a triangle into three equal areas is the Centroid.
Centroid $P$ : $(\frac{17}{6}, \frac{8}{3})$
Final Distance $P Q$ : $5$ units

00:00 / 00:00

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

Solution Diagram

Analyzing the Setup

Welcome to a beautiful exploration of coordinate geometry. We are given the vertices of triangle

A B C

A (1, 0)

B (6, 2)

, and

C (\frac{3}{2}, 6)

We are looking for a point

P

inside the triangle such that the areas of

△ A P C

△ A P B

, and

△ B P C

are equal.

In any triangle, the unique point that partitions the triangle into three regions of equal area is the centroid. This point acts as the center of mass, ensuring the geometric balance of the figure.

The Coordinate Hunt

Since

P

is the centroid, we calculate its coordinates using the arithmetic mean of the vertices' coordinates. The formula for the centroid

P (x_{P}, y_{P})

is given by:

P = (\frac{x _{A} + x _{B} + x _{C}}{3}, \frac{y _{A} + y _{B} + y _{C}}{3})

For the

x

-coordinate, we calculate:

x_{P} = \frac{1 + 6 + \frac{3}{2}}{3} = \frac{7 + \frac{3}{2}}{3} = \frac{\frac{17}{2}}{3} = \frac{17}{6}

For the

y

-coordinate, we calculate:

y_{P} = \frac{0 + 2 + 6}{3} = \frac{8}{3}

Thus, the coordinates of our point

P

are

(\frac{17}{6}, \frac{8}{3})

The Final Bridge

We are given a second point

Q (- \frac{7}{6}, - \frac{1}{3})

and must find the distance

P Q

. We apply the distance formula:

P Q = (x_{P} - x_{Q})^{2} + (y_{P} - y_{Q})^{2}

Substituting our known values into the equation:

P Q = (\frac{17}{6} - (- \frac{7}{6}))^{2} + (\frac{8}{3} - (- \frac{1}{3}))^{2}

Simplifying the differences within the parentheses:

P Q = (\frac{17 + 7}{6})^{2} + (\frac{8 + 1}{3})^{2} = (\frac{24}{6})^{2} + (\frac{9}{3})^{2}

This reduces to:

P Q = 4^{2} + 3^{2} = 16 + 9 = 25

The final distance is $5$ units.

Common Traps & FAQs

Wait, why is the point

P

the centroid? Could it be the incenter or orthocenter?

That is a fantastic question! In any triangle, the centroid is the unique point that divides the triangle into three smaller triangles of equal area. If you connect the centroid to the three vertices, you create three triangles, each with an area exactly one-third of the total area of the original triangle. The incenter or orthocenter do not possess this property.

Do I really need to memorize the centroid formula, or can I derive it?

You absolutely can derive it! The centroid is the intersection of the medians. You can find the midpoint of one side, connect it to the opposite vertex to form a median, and then find the point that divides that median in a

2 : 1

ratio. However, for competitive exams like JEE, memorizing the formula

P = (\frac{x _{1} + x _{2} + x _{3}}{3}, \frac{y _{1} + y _{2} + y _{3}}{3})

saves precious time.

Is there a trick to the distance calculation? The fractions look intimidating.

I understand the hesitation! When you see fractions like

\frac{17}{6}

and

- \frac{7}{6}

, it might look messy. But notice that when you subtract them, the denominators are the same, which makes the arithmetic much cleaner. Always look for that simplification before you start squaring large numbers. It is often a sign that you are on the right track!

Enter Numerical Value:

Visualized Solution

Visualizing Triangle $A B C$

Identifying Point $P$

Applying the Centroid Formula

Calculating $x$ -coordinate of $P$

Calculating $y$ -coordinate of $P$

Defining Point $Q$ and Distance Formula

Substituting Values in Distance Formula

Simplifying the Coordinate Differences

Final Calculation

Summary and Key Takeaway

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

Analyzing the Setup

The Coordinate Hunt

The Final Bridge

Similar Questions

Let $O (0, 0), P (3, 4), Q (6, 0)$ be the vertices of the triangles $O P Q$ . The point $R$ inside the triangle $O P Q$ is such that the triangles $O P R, P QR, O QR$ are of equal area. The coordinates of $R$ are

Let $A B C$ be the triangle such that the equations of lines $A B$ and $A C$ be $3 y - x = 2$ and $x + y = 2$ , respectively, and the points $B$ and $C$ lie on $x$ -axis. If $P$ is the orthocentre of the triangle $A B C$ , then the area of the triangle $P B C$ is equal to

The equations of the sides AB, BC and CA of a triangle ABC are: $2 x + y = 0$ , $x + p y = 21 a$ , $(a \neq = 0)$ and $x - y = 3$ respectively. Let $P (2, a)$ be the centroid of $Δ A B C$ . Then $(B C)^{2}$ is equal to

Let the circumcentre of a triangle with vertices $A (a, 3), B (b, 5)$ and $C (a, b), ab > 0$ be $P (1, 1)$ . If the line $A P$ intersects the line $B C$ at the point $Q (k_{1}, k_{2})$ , then $k_{1} + k_{2}$ is equal to :

Let the area of a $△ P QR$ with vertices $P (5, 4)$ , $Q (- 2, 4)$ and $R (a, b)$ be 35 square units. If its orthocenter and centroid are $O (2, \frac{14}{5})$ and $C (c, d)$ respectively, then $c + 2 d$ is equal to

In a triangle ABC, coordinates of A are (1, 2) and the equations of the medians through B and C are respectively, $x + y = 5$ and $x = 4$ . Then area of $Δ A B C$ (in sq. units) is:

If $P (6, 1)$ be the orthocentre of the triangle whose vertices are $A (5, - 2)$ , $B (8, 3)$ and $C (h, k)$ , then the point $C$ lies on the circle:

If a vertex of a triangle is $(1, 1)$ and the mid points of two sides through this vertex are $(- 1, 2)$ and $(3, 2)$ then the centroid of the triangle is

In an isosceles triangle $A B C$ , the vertex $A$ is $(6, 1)$ and the equation of the base $B C$ is $2 x + y = 4$ . Let the point $B$ lie on the line $x + 3 y = 7$ . If $(α, β)$ is the centroid $Δ A B C$ , then $15 (α + β)$ is equal to :

Enter Numerical Value:

Visualized Solution

Visualizing Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABC

Identifying Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Applying the Centroid Formula

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-coordinate of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-coordinate of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

Defining Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q and Distance Formula

Substituting Values in Distance Formula

Simplifying the Coordinate Differences

Final Calculation

Summary and Key Takeaway

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

Analyzing the Setup

The Coordinate Hunt

The Final Bridge

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In a triangle ABC, coordinates of A are (1, 2) and the equations of the medians through B and C are respectively, x+y=5 and x=4. Then area of ΔABC (in sq. units) is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If P(6,1) be the orthocentre of the triangle whose vertices are A(5,−2), B(8,3) and C(h,k), then the point C lies on the circle:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a vertex of a triangle is (1,1) and the mid points of two sides through this vertex are (−1,2) and (3,2) then the centroid of the triangle is

Visualizing Triangle $A B C$

Identifying Point $P$

Calculating $x$ -coordinate of $P$

Calculating $y$ -coordinate of $P$

Defining Point $Q$ and Distance Formula

In a triangle ABC, coordinates of A are (1, 2) and the equations of the medians through B and C are respectively, $x + y = 5$ and $x = 4$ . Then area of $Δ A B C$ (in sq. units) is:

If $P (6, 1)$ be the orthocentre of the triangle whose vertices are $A (5, - 2)$ , $B (8, 3)$ and $C (h, k)$ , then the point $C$ lies on the circle:

If a vertex of a triangle is $(1, 1)$ and the mid points of two sides through this vertex are $(- 1, 2)$ and $(3, 2)$ then the centroid of the triangle is