Question

30 March 20201. The straight line L1 passes through the point A (5,1) and is perpendicular to the line L2 whose equation is 3x - y = 4.Find (i) the equation of L1 (ii) the coordinates of the point of intersection of L1 and L2 (iii) the perpendicular distance from the point A (5,1) to the line L2

Answer 1

A line perpendicular to 3x - y = 4 must have the form
x + 3y = C
plug in the point (5,1) ---> 5 + 3 = C = 8

new line: x + 3y = 8

intersection: from the first, y = 3x-4
plug into 2nd,
x + 3(3x-4) = 8
x+9x-12 = 8
10x = 20
x = 2 , then y = 2 <----- intersect at (2,2)

distance between (2,2) and (5,1)
= √(3^2 + (-1)^2) = √10

check using distance from a point to a line formula
line: 3x-y-4=0 to point (5,1)
= |3(5) - 1 - 4|/√(1+9)
= 10/√10 = √10

Answer 2

L2: slope = -A/B = -3/-1 = 3.
(I). L1: A(5, 1), slope = -(1/3) = -1/3.
Y = mx + b
1 = (-1/3)5 + b
b = 8/3.
Y = -x/3 + 8/3
x + 3y = 8.

(ii). Multiply Eq1 by 3 and add:
9x - 3y = 12
x + 3y = 8
sum: 10x = 20. X = 2.
2 + 3y = 8. Y=2.

(iii). (2, 2), (5, 1).
d^2 = (5-2)^2 + (1-2)^2 = 10. d = 3.2.