Question

Two fixed points A(-2, 3) and B(5, 3) are given. Find the equation of locus of a moving point P such that area of triangle PAB is 20 sq units. *No Spam* ✖✖

Answer 1

Given:

Two fixed points-

A(-2, 3) and B(5,3)

To Find:

Equation of locus of given moving point P  such that area of triangle PAB is 20 sq units

Solution:

We know that,

• For a triangle having coordinates of vertices as (x₁,y₁), (x₂,y₂) and (x₃,y₃)

$\pink{\underline{\boxed{\sf{Area\:of\:\triangle=\Bigg|\dfrac{1}{2}\Big(x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\Big)\Bigg|}}}}$

• If |x|= c, then x= ±c

━━━━━━━━━━━━━━━━━━━━━

Let the coordinates of movable point P be (h,k) and the area of triangle PAB be Δ

Now, in triangle PAB

Here,

P(h,k)⟼ (x₁,y₁)

A(-2,3)⟼ (x₂,y₂)

B(5,3)⟼ (x₃,y₃)

So,

$\longrightarrow\rm{\triangle=\dfrac{1}{2}\Bigg|\Big(h(3-3)+(-2)(3-k)+5(k-3)\Big)\Bigg|}$

$\longrightarrow\rm{20=\dfrac{1}{2}\Bigg|\Big(h(0)-2(3-k)+5(k-3)\Big)\Bigg|}$

$\longrightarrow\rm{\dfrac{1}{2}\Bigg|\Big(h(0)-2(3-k)+5(k-3)\Big)\Bigg|=20}$

$\longrightarrow\rm{\Bigg|\Big(h(0)-2(3-k)+5(k-3)\Big)\Bigg|=20\times2}$

$\longrightarrow\rm{\Bigg|\Big(h(0)-2(3-k)+5(k-3)\Big)\Bigg|=40}$

$\longrightarrow\rm{\Bigg|-6+2k+5k-15\Bigg|=40}$

$\longrightarrow\rm{\Bigg|7k-21\Bigg|=40}$

$\longrightarrow\rm{7k-21=\pm40}$

On substituting (h,k) with (x,y), we get

$\longrightarrow\rm{7y-21=\pm40}$

━━━━━━━━━━━━━━━━━━━━━

Now, we have to consider two cases

Case 1:- When 7y-21=40

$\longrightarrow\rm{7y-21=40}$

$\longrightarrow\rm{7y=40+21}$

$\longrightarrow\rm{7y=61}$

$\longrightarrow\rm\green{y=\dfrac{61}{7}}$

Case 2:- When 7y-21= -40

$\longrightarrow\rm{7y-21=-40}$

$\longrightarrow\rm{7y=-40+21}$

$\longrightarrow\rm{7y=-19}$

$\longrightarrow\rm\green{y=\dfrac{-19}{7}}$

Hence, the locus of moving point P is y=61/7 and y= -19/7.

Answer 2

$\LARGE{\red{\boxed{\purple{\underline{\orange{\mathtt{Question:↓}}}}}}}$

Two fixed points A(-2, 3) and B(5, 3) are given. Find the equation of locus of a moving point P such that area of triangle PAB is 20 sq units.

$\LARGE{\red{\boxed{\purple{\underline{\orange{\mathtt{Given:↓}}}}}}}$

Two fixed points-

A(-2, 3) and B(5,3)

$\LARGE{\red{\boxed{\purple{\underline{\orange{\mathtt{To\: Find:↓}}}}}}}$

Equation of locus of given moving point P such that area of triangle PAB is 20 sq units

$\LARGE{\red{\boxed{\purple{\underline{\orange{\mathtt{Solution:↓}}}}}}}$

We know that,

For a triangle having coordinates of vertices as (x₁,y₁), (x₂,y₂) and (x₃,y₃)

$\pink{\underline{\boxed{\sf{Area\:of\:\triangle=\Bigg|\dfrac{1}{2}\Big(x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\Big)\Bigg|}}}}$

Area of △=

If |x|= c, then x= ±c

━━━━━━━━━━━━━━━━━━━━━

Let the coordinates of movable point P be (h,k) and the area of triangle PAB be Δ

Now, in triangle PAB

Here,

P(h,k)⟼ (x₁,y₁)

A(-2,3)⟼ (x₂,y₂)

B(5,3)⟼ (x₃,y₃)

So,

$\longrightarrow\rm{\triangle=\dfrac{1}{2}\Bigg|\Big(h(3-3)+(-2)(3-k)+5(k-3)\Big)\Bigg|}⟶△=$

$\longrightarrow\rm{20=\dfrac{1}{2}\Bigg|\Big(h(0)-2(3-k)+5(k-3)\Big)\Bigg|}⟶20=$

$\longrightarrow\rm{\dfrac{1}{2}\Bigg|\Big(h(0)-2(3-k)+5(k-3)\Big)\Bigg|=20}⟶$

$\longrightarrow\rm{\Bigg|\Big(h(0)-2(3-k)+5(k-3)\Big)\Bigg|=20\times2}⟶$

$\longrightarrow\rm{\Bigg|\Big(h(0)-2(3-k)+5(k-3)\Big)\Bigg|=40}⟶$

$\longrightarrow\rm{\Bigg|-6+2k+5k-15\Bigg|=40}⟶$

$\longrightarrow\rm{\Bigg|7k-21\Bigg|=40}⟶$

$\longrightarrow\rm{7k-21=\pm40}⟶7k−21=±40$

On substituting (h,k) with (x,y), we get

$\longrightarrow\rm{7y-21=\pm40}⟶7y−21=±40$

━━━━━━━━━━━━━━━━━━━━━

Now, we have to consider two cases

Case 1:-

When 7y-21=40

$\longrightarrow\rm{7y-21=40}⟶7y−21=40$

$\longrightarrow\rm{7y=40+21}⟶7y=40+21$

$\longrightarrow\rm{7y=61}⟶7y=61$

$\longrightarrow\rm\green{y=\dfrac{61}{7}}$

Case 2:-

When 7y-21= -40

$\longrightarrow\rm{7y-21=-40}⟶7y−21=−40$

$\longrightarrow\rm{7y=-40+21}⟶7y=−40+21$

$\longrightarrow\rm{7y=-19}⟶7y=−19$

$\longrightarrow\rm\green{y=\dfrac{-19}{7}}$

Hence, the locus of moving point P is y=61/7 and y= -19/7.