P x - y - A 3-4 and B 5-2 are the vertices of triangle PAB such that - PA - PB - and area of - P AB -10 sq- units- then P A - k -5 units- Find thevalue of k-A- 2B- -5C- 1D- 3

The correct option is A 2 Given, |PA|=|PB| ∴ PA^2=PB^2 ⇒ (x 3)^2+(y 4)^2=(x 5)^2+(y+2)^2 ⇒ x^2 6x+9+y^2 8y+16=x^2 10x+25+y^2+4y+4 ⇒ 4x 12y=4 ⇒ x 3y=1 … (i ...

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Byju's Answer

Standard IX

Mathematics

Area of Triangle Given Its Vertices

P x , y , A 3...

Question

P(x, y), A(3, 4) and B (5, -2) are the vertices of triangle PAB such that |PA|=|PB| and area of $Δ P A B = 10$ sq. units, then $P A = k \sqrt{5}$ units. Find the value of k.

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Solution

The correct option is A 2

Given, |PA|=|PB|
$∴ P A^{2} = P B^{2} \Rightarrow (x - 3)^{2} + (y - 4)^{2} = (x - 5)^{2} + (y + 2)^{2} \Rightarrow x^{2} - 6 x + 9 + y^{2} - 8 y + 16 = x^{2} - 10 x + 25 + y^{2} + 4 y + 4 \Rightarrow 4 x - 12 y = 4 \Rightarrow x - 3 y = 1 \dots (i)$
area of $Δ P A B = \frac{1}{2} | x (4 + 2) + 3 (- 2 - y) + 5 (y - 4) | \Rightarrow 10 = \frac{1}{2} | 6 x - 6 - 3 y + 5 y - 20 | \Rightarrow 20 = 6 x + 2 y - 26 \Rightarrow 6 x + 2 y - 26 = 20 o r 6 x + 2 y - 26 = - 20 \Rightarrow 6 x + 2 y = 46 o r 6 x + 2 y = 6 3 x + y = 23 \dots (2)$

or $3 x + y = 3 \dots (3)$
Solving (1) and (3), we get
$x = 1, y = 0$
Solving (1) and (2), we get
$x = 7, y = 2$
$∴$ Coordinates of P are (7, 2) and (1, 0)
Length of $P A = \sqrt{(7 - 3)^{2} + (2 - 4)^{2}} = \sqrt{16 + 4} = 2 \sqrt{5} ∴ 2 \sqrt{5} = k \sqrt{5} \Rightarrow k = 2$
Also, Length of $P A = \sqrt{(1 - 3)^{2} + (0 - 4)^{2}} = \sqrt{4 + 16} = 2 \sqrt{5} ∴ 2 \sqrt{5} = k \sqrt{5} \Rightarrow k = 2$

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P(x, y), A(3, 4) and B (5, -2) are the vertices of triangle PAB such that |PA|=|PB| and area of ΔPAB=10 sq. units, then PA=k√5 units. Find the value of k.

P(x, y), A(3, 4) and B (5, -2) are the vertices of triangle PAB such that |PA|=|PB| and area of $Δ P A B = 10$ sq. units, then $P A = k \sqrt{5}$ units. Find the value of k.