Mathematics 2011 JAMB Past Questions - Question 33: Find the equation of a line perpendicular to line 2y=5x+4 which passes through (4,2).

Correct Answer

B

Explanation

2y = 5x +4 (4,2)therefore y = 5x/2 +4 comparing withy = mx +etherefore m = 5/2since they are perpendicularm1 m2 = -1m2 = -1/m15/2 = -1x2/5= - 2/5the equator of the line is thusy = mn+c (4,2)2 = -2/5 (4) + c2/1 +8/5 = ctherefore c = 18/510+8/5 = ctherefore y = -2/5 x +18/55y = -2x +18or 5y +2x - 18 = 0To find the equation of a line perpendicular to the line 2y = 5x + 4, we need to determine the slope of the given line and then find the negative reciprocal of that slope.The given line is in the form y = mx + b, where m is the slope. Comparing the given equation to this form, we can see that the slope of the given line is 5/2.The negative reciprocal of 5/2 is -2/5. This will be the slope of the line perpendicular to the given line.Now, we can use the point-slope form of a linear equation to find the equation of the line perpendicular to the given line that passes through the point (4, 2).The point-slope form is given by y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line and m is the slope.Plugging in the values, we have:y - 2 = (-2/5)(x - 4)Simplifying, we get:y - 2 = (-2/5)x + 8/5Adding 2 to both sides, we have:y = (-2/5)x + 18/5Therefore, the equation of the line perpendicular to 2y = 5x + 4 that passes through (4, 2) is y = (-2/5)x + 18/5.