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Mathematics

Find the slope of the line perpendicular to AB if :

(i) A = (0, -5) and B = (-2, 4)

(ii) A = (3, -2) and B = (-1, 2)

Straight Line Eq

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Answer

(i) A = (0, -5) and B = (-2, 4)

By formula,

Slope (m) = y2−y1x2−x1\dfrac{y2 - y1}{x2 - x1}

Substituting values we get,

 Slope of AB(m1)=4−(−5)−2−0=4+5−2=9−2=−92.\text{ Slope of AB}(m_1) = \dfrac{4 - (-5)}{-2 - 0} \\[1em] = \dfrac{4 + 5}{-2} \\[1em] = \dfrac{9}{-2} = -\dfrac{9}{2}.

Let m2 be the slope of perpendicular line.

We know that,

Product of slope of perpendicular lines = -1.

∴ m1.m2 = -1

⇒−92×m2=−1⇒m2=−1×−29⇒m2=29.\Rightarrow -\dfrac{9}{2} \times m2 = -1 \\[1em] \Rightarrow m2 = -1 \times -\dfrac{2}{9} \\[1em] \Rightarrow m_2 = \dfrac{2}{9}.

Hence, slope of the line perpendicular to AB = 29.\dfrac{2}{9}.

(ii) A = (3, -2) and B = (-1, 2)

By formula,

Slope (m) = y2−y1x2−x1\dfrac{y2 - y1}{x2 - x1}

Substituting values we get,

 Slope of AB(m1)=2−(−2)−1−3=4−4=−1.\text{ Slope of AB}(m_1) = \dfrac{2 - (-2)}{-1 - 3} \\[1em] = \dfrac{4}{-4} \\[1em] = -1.

Let m2 be the slope of perpendicular line.

We know that,

Product of slope of perpendicular lines = -1.

∴ m1.m2 = -1

⇒−1×m2=−1⇒m2=−1−1⇒m2=1.\Rightarrow -1 \times m2 = -1 \\[1em] \Rightarrow m2 = \dfrac{-1}{-1} \\[1em] \Rightarrow m_2 = 1.

Hence, slope of the line perpendicular to AB = 1.

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