Let the coordinates of P - Q and R be 6-3-3-5 and 4-2 respectively- If A - x - y - then the ratio of area of AQR to PQR is A- 2-7 if x -0- y -0B- 2-7 if x -1- y -2C- 1 if x -5- y -4D- 1 if x -4- y -5

The correct options are A 27 if x = 0, y = 0 C 1 if x = 5, y = 4 D 1 if x = 4, y = 5Given : P nbsp; = (6, 3), Q = ( 3, 5), R = (4, 2) nbsp; Now, nbsp;Ar ...

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Byju's Answer

Standard XII

Mathematics

Area Method to Find Condition for Co-Linearity

Let the coord...

Question

Let the coordinates of $P, Q$ and $R$ be $(6, 3),$ $(- 3, 5)$ and $(4, - 2)$ respectively. If $A \equiv (x, y),$ then the ratio of area of $△ A Q R$ to $△ P Q R$ is

\frac{2}{7}

x = 0, y = 0

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\frac{2}{7}

x = 1, y = 2

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1

x = 5, y = 4

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1

x = 4, y = 5

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Solution

The correct options are
A $\frac{2}{7}$ if $x = 0, y = 0$
C $1$ if $x = 5, y = 4$
D $1$ if $x = 4, y = 5$
Given : $P = (6, 3), Q = (- 3, 5), R = (4, - 2)$
Now, $A r e a o f Δ P Q R$
$= \frac{1}{2} ∣ ∣ ∣ \begin{matrix} x_{1} & x_{2} & x_{3} & x_{1} y_{1} & y_{2} & y_{3} & y_{1} \end{matrix} ∣ ∣ ∣ = \frac{1}{2} ∣ ∣ ∣ \begin{matrix} 6 & - 3 & 4 & 6 3 & 5 & - 2 & 3 \end{matrix} ∣ ∣ ∣ = \frac{1}{2} | (30 + 9) + (6 - 20) + (12 + 12) | = \frac{49}{2}$

Also, $A r e a o f Δ A Q R$
$= \frac{1}{2} ∣ ∣ ∣ \begin{matrix} x & - 3 & 4 & x y & 5 & - 2 & y \end{matrix} ∣ ∣ ∣ = \frac{1}{2} | (5 x + 3 y) + (6 - 20) + (4 y + 2 x) | = \frac{| 7 x + 7 y - 14 |}{2}$

Now, $\frac{A r e a o f Δ A Q R}{A r e a o f Δ P Q R} = \frac{| x + y - 2 |}{7}$

Therefore, when $x = 0, y = 0$ , we get
$\frac{A r e a o f Δ A Q R}{A r e a o f Δ P Q R} = \frac{2}{7}$

When $x = 1, y = 2$ , we get
$\frac{A r e a o f Δ A Q R}{A r e a o f Δ P Q R} = \frac{1}{7}$

When $x = 5, y = 4$ , we get
$\frac{A r e a o f Δ A Q R}{A r e a o f Δ P Q R} = 1$

When $x = 4, y = 5$ , we get
$\frac{A r e a o f Δ A Q R}{A r e a o f Δ P Q R} = 1$

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Similar questions

Q. Let the coordinates of

P, Q

and

R

(6, 3),

(- 3, 5)

and

(4, - 2)

respectively. If

A \equiv (x, y),

then the ratio of area of

△ A Q R

△ P Q R

Q. If P = 7x² + 5xy − 9y², Q = 4y² − 3x² − 6xy and R = −4x² + xy + 5y², show that P + Q + R = 0.

Q. The coordinates of

A, B, C

are

(6, 3), (- 3, 5), (4, - 2)

, respectively, and

P

is any point

(x, y)

. Show that the ratio of the area of

△ P B C

to that of

△ A B C

| x + y - 2 | / 7

Find the values of $x$ and $y$ if :

$\frac{5}{2 + x} + \frac{1}{y - 4} = 2$

$\frac{6}{2 + x} - \frac{3}{y - 4} = 1$

Here, x $\neq$ -2 and y $\neq$ 4.

Q. If

P = 7 x^{2} + 5 x y - 9 y^{2}, Q = 4 y^{2} - 3 x^{2}

and

R = 4 x^{2} + x y + 5 y^{2}

show that

P + Q + R = 8 x^{2} + 6 x y

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Let the coordinates of P,Q and R be (6,3), (−3,5) and (4,−2) respectively. If A≡(x,y), then the ratio of area of △AQR to △PQR is

Let the coordinates of $P, Q$ and $R$ be $(6, 3),$ $(- 3, 5)$ and $(4, - 2)$ respectively. If $A \equiv (x, y),$ then the ratio of area of $△ A Q R$ to $△ P Q R$ is