find all the points Find all of the points on the line y 2x

Find all of the points on the line y = 2x + 6 which are 4 units from the point (-1, 8). (x, y) = () (smaller x-value) (x, y) = () (larger x-value)

Solution

Find all the points on the line y = 2x + 6 which are 4 units from the point ( -1, 8)

The distance d, between two pints P_{1 :} (x₁, y₁) and P₂ (x₂ , y₂) = [ ( (x₁ – x₂)² + ( y₁ – y₂)²] . We are to determine all the points (x, y), which are on the line y = 2x + 6 and are 4 units away from the point ( -1, 8). Let P: (x, y) be such a point. Since P lies on the line y = 2x + 6 …(1). Also, since the distance of P from the point ( -1, 8) is 4 units, we have { ( x+1)² + ( y - 8)²] = 4. On taking squares of both the sides, we have ( x+1)² + ( y - 8)² = 16 …(2). Now, from the 1^st equation, we have y = 2x + 6. On substituting this value of y in the 2^nd equation, we have (x + 1)² + ( 2x + 6 - 8)² = 16 or, (x + 1)² + ( 2x - 2)² = 16 or, x² + 2x +1 + 4x² - 8x + 4 = 16 or, 5x² - 6x - 11 = 0…(3) Therefore x = [- (-6) ± { ( -6)² – 4(5) (-11)}]/ 2*5 or, x = [ 6 ± ( 36 + 220) ] / 10 or, x = ( 6 ± 256)/ 10 or, x = (6 ± 16)/10. Therefore, either x = 22/10 = 11/5 or x = -1. When x = -1, y = 2(-1) + 6 = - 2 + 6 = 4; when x = 11/5, we have y = 2(11/5) + 6 = 52/5. Thus the two points ( -1, 4) ( smaller value) and ( 11/5 , 52/5 ) ( larger value) are on the line y = 2x + 6 and are 4 units distance from the point ( -1, 8)