Mathematics 2000 JAMB Past Questions - Question 28: 3y=4x-1 and Ky=x+3 are equations of two straight lines .if the two lines are perpendicular to each other .find k

Correct Answer

A

Explanation

3y = 4x -1 = y = 4/3 x -1 m1 = 4/3ky = x +3 = y 1/k x + 3/k msup2 = 1/kfor two striaght lines to be perpendicular to each other, msup1 msup2 = -1therefore 4/3 x 1/k = -1 4 = -3ktherefore k = -4 /3To find the value of k such that the two lines 3y = 4x - 1 and Ky = x + 3 are perpendicular to each other, we can use the property that the product of the slopes of two perpendicular lines is -1.First, let's rewrite the equations in slope-intercept form (y = mx + b), where m is the slope:For the equation 3y = 4x - 1:Dividing both sides by 3, we get y = (4/3)x - 1/3.So, the slope of this line is m1 = 4/3.For the equation Ky = x + 3:Dividing both sides by K, we get y = (1/K)x + 3/K.So, the slope of this line is m2 = 1/K.Since the two lines are perpendicular, we have the relationship:m1 * m2 = -1(4/3) * (1/K) = -1To solve for K, we can multiply both sides by 3K:4 = -3KDividing both sides by -3, we get:K = -4/3Therefore, the value of K that makes the two lines perpendicular to each other is K = -4/3.