Let A(–1, 1) and B(2, 3) be two points and P be a variable point above the line AB such that the area of is 10. If the locus of P is ax + by = 15, then 5a + 2b is: [2024]

Question

Let A(–1, 1) and B(2, 3) be two points and P be a variable point above the line AB such that the area of $△ P A B$ is 10. If the locus of P is ax + by = 15, then 5a + 2b is: [2024]

Smart Achievers · Accepted Answer

(4)

Let coordinates of P be (h, k).

Area of $△ P A B$ = 10sq. units

$\Rightarrow 10 = \frac{1}{2}$ $| \begin{matrix} h \\ - 1 \\ 2 \end{matrix} \begin{matrix} k \\ 1 \\ 3 \end{matrix} \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} |$

$\Rightarrow \pm 20 = h (- 2) - k (- 3) + 1 (- 5)$

$\Rightarrow - 2 h + 3 k = 25$

( $∵$ P is above the line AB, so we take only +ve value)

$\Rightarrow \frac{- 2 \times 3}{5} h + \frac{3 \times 3}{5} k = \frac{3 \times 25}{5}$

$\Rightarrow \frac{- 6}{5} h + \frac{9}{5} k = 15$

On comparing it with ax + by = 15, we get $a = \frac{- 6}{5}$ and $b = \frac{9}{5}$

. $5 a + 2 b = - 6 + \frac{18}{5} = \frac{- 12}{5}$

Solution

Q.
Let A(–1, 1) and B(2, 3) be two points and P be a variable point above the line AB such that the area of $△ P A B$ is 10. If the locus of P is ax + by = 15, then 5a + 2b is: [2024]

1 6

2 4

3 $- \frac{6}{5}$

4 $- \frac{12}{5}$

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Solution

Q. Let A(–1, 1) and B(2, 3) be two points and P be a variable point above the line AB such that the area of △PAB is 10. If the locus of P is ax + by = 15, then 5a + 2b is: [2024]

1 6

2 4

3 - 65

4 - 125

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Q.
Let A(–1, 1) and B(2, 3) be two points and P be a variable point above the line AB such that the area of $△ P A B$ is 10. If the locus of P is ax + by = 15, then 5a + 2b is: [2024]

3 $- \frac{6}{5}$

4 $- \frac{12}{5}$