What is the perpendicular distance of a point 2... - JAMB Mathematics 1992 Question
What is the perpendicular distance of a point (2, 3) from the line 2x - 4y + 3 = 0?
A
\(\frac{\sqrt{5}}{2}\)
B
\(\frac{\sqrt{5}}{20}\)
C
\(\frac{5}{\sqrt{13}}\)
D
6
correct option: a
2x - 4y + 3 = 0
Required distance = \(\frac{(2 \times 2) + 3(-4) + 3}{\sqrt{2^2} + (-4)^2}\)
= \(\frac{4 - 12 + 3}{\sqrt{20}}\)
= \(\frac{-5}{-2\sqrt{5}}\}
= -\{\frac{\sqrt{5}{2}\)
Required distance = \(\frac{(2 \times 2) + 3(-4) + 3}{\sqrt{2^2} + (-4)^2}\)
= \(\frac{4 - 12 + 3}{\sqrt{20}}\)
= \(\frac{-5}{-2\sqrt{5}}\}
= -\{\frac{\sqrt{5}{2}\)
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