Ask Singapore Homework?
Upload a photo of a Singapore homework and someone will email you the solution for free.
Question
secondary 3 | A Maths
One Answer Below
Anyone can contribute an answer, even non-tutors.
asap
Date Posted: 3 years ago
Views: 183
See 1 Answer
done
{{ upvoteCount }} Upvotes
clear
{{ downvoteCount * -1 }} Downvotes
Since line L is perpendicular to line AB, you can find the gradient of line L by dividing -1 to the gradient of line AB. ( -1 ÷ 2/3)
*refer to gradient of normal line
Since line L is the perpendicular bisector of line AB, this means that point P is the midpoint of coordinates A and B. Hence, you can find the coordinates of P.
Afterwards, you can construct your equation for line L using the formula y1 - y2 = m ( x1 - x2 )
Since point C is directly above A, the x-coordinates for both C and A are the same ( -1 )
*refer to gradient of normal line
Since line L is the perpendicular bisector of line AB, this means that point P is the midpoint of coordinates A and B. Hence, you can find the coordinates of P.
Afterwards, you can construct your equation for line L using the formula y1 - y2 = m ( x1 - x2 )
Since point C is directly above A, the x-coordinates for both C and A are the same ( -1 )
Date Posted:
3 years ago
360 Tutoring Program