The angle between the line passing through the points (1,-2) (3,2) and the line x+2y-7=0 is

\frac{π}{1}

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\frac{π}{2}

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\frac{π}{3}

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\frac{π}{4}

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Solution

The correct option is A $\frac{π}{1}$
$L_1: (1, -2), (3, 2)
$(y + 2) = \frac{2 + 2}{3 - 1} (x - 1)$
$\Rightarrow (y + 2) = 2 (x - 1)$
$\Rightarrow 2 x - y - 3 = 0$ , $m_{1} = 2$
$L_{2} : x + 2 y - 7 = 0$ , $m = \frac{- 1}{2}$
$tan θ = \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}} = \frac{2 + \frac{1}{2}}{1 + 2 (- \frac{1}{2})} = \frac{4 + 1}{\frac{2}{0}} = \frac{1}{0} = \infty$
$∴ tan θ = tan π$
$∴ θ = π$ .

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The angle between the line passing through the points (1,-2) (3,2) and the line x+2y-7=0 is

The correct option is A π1$L_1: (1, -2), (3, 2)(y+2)=2+23−1(x−1)⇒(y+2)=2(x−1)⇒2x−y−3=0, m1=2L2:x+2y−7=0, m=−12tanθ=m1−m21+m1m2=2+121+2(−12)=4+120=10=∞∴tanθ=tanπ∴θ=π.