JAMB - Mathematics (2000 - No. 21)
3y = 4x - 1 and Ky = x + 3 are equations of two straight lines. If the two lines are perpendicular to each other, find K.
-\(\frac{4}{3}\)
-\(\frac{3}{4}\)
\(\frac{3}{4}\)
\(\frac{4}{3}\)
ব্যাখ্যা
Grad of 3y = 4x - 1
y = \(\frac{4x}{3}\) - \(\frac{1}{3}\)
Grad = \(\frac{4}{3}\)
Grad of Ky = x + 3
y = \(\frac{\text{x}}{\text{k}}\) + \(\frac{3}{4}\)
Grad = \(\frac{1}{\text{k}}\)
Since the two lines are perpendicular,
\(\frac{1}{\text{k}}\) = -\(\frac{3}{4}\)
-3k = 4
k = -\(\frac{4}{3}\)
y = \(\frac{4x}{3}\) - \(\frac{1}{3}\)
Grad = \(\frac{4}{3}\)
Grad of Ky = x + 3
y = \(\frac{\text{x}}{\text{k}}\) + \(\frac{3}{4}\)
Grad = \(\frac{1}{\text{k}}\)
Since the two lines are perpendicular,
\(\frac{1}{\text{k}}\) = -\(\frac{3}{4}\)
-3k = 4
k = -\(\frac{4}{3}\)