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Slope of a Line Connecting Two Points

Line passing ...

Question

Line passing through the points $2 \to a + 3 \to b - \to c, 3 \to a + 4 \to b - 2 \to c$ intersects the line through the points $\to a - 2 \to b + 3 \to c, \to a - 6 \to b + 6 \to c$ at $P$ . Then position vector of $P$ is

A

2 \to a + \to b

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B

\to a + 2 \to b

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C

3 \to a + 4 \to b

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D

2 \to a - \to b

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Solution

The correct option is B $\to a + 2 \to b$
Let position vectors $- - \to O A = 2 \to a + 3 \to b - \to c,$
$- - \to O B = 3 \to a + 4 \to b - 2 \to c, - - \to O C = \to a - 2 \to b + 3 \to c, - - \to O D = \to a - 6 \to b + 6 \to c$
Equation of line through the points $A$ and $B$ is $\to r = - - \to O A + λ - - \to A B$
$\Rightarrow \to r = (2 \to a + 3 \to b - \to c) + λ (\to a + \to b - \to c) \dots (i)$ is the first line.
Equation of line through the points $C$ and $D$ is $\to r = - - \to O C + μ - - \to C D$
$\Rightarrow \to r = (\to a - 2 \to b + 3 \to c) + μ (- 4 \to b + 3 \to c) \dots (i i)$ is the second line.
For point of intersection $P$
$(i) = (i i)$
$\Rightarrow (2 \to a + 3 \to b - \to c) + λ (\to a + \to b - \to c) = (\to a - 2 \to b + 3 \to c) + μ (- 4 \to b + 3 \to c)$
$(2 + λ) \to a + (3 + λ) \to b + (- 1 - λ) \to c = (1) \to a + (- 2 - 4 μ) \to b + (3 + 3 μ) \to c$
On comparing
$2 + λ = 1 \Rightarrow λ = - 1$
putting in eqution $(i)$
$P = (2 \to a + 3 \to b - \to c) + (- 1) (\to a + \to b - \to c)$
Hence $P = \to a + 2 \to b$

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Slope of a Line Connecting Two Points

Standard XII Mathematics

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