Han Zhi Yuan's answer to Jason Lim's Secondary 3 A Maths Singapore question.
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There is no mention of point D in the question so I'm going to assume CS = DS to find the coordinates of D.
The gradient of line AB is -1.
y = mx + c. A = (1,8)
When x = 1, 8 = (-1)1 + c. So c = 9.
Equation of line AB is y = 9 - x.
Point S falls on both lines y = 9 - x and y = x + 5.
=> 9 - x = x + 5
=> x = 2.
When x = 2, y = 7. This means S = (2,7).
We already know C = (5,10).
From (5,10) to (2,7), both x and y-coordinates goes down by 3. If we go down again, we get (-1,4) which is point D.
Hence coordinates of point D = (-1,4).
The gradient of line AB is -1.
y = mx + c. A = (1,8)
When x = 1, 8 = (-1)1 + c. So c = 9.
Equation of line AB is y = 9 - x.
Point S falls on both lines y = 9 - x and y = x + 5.
=> 9 - x = x + 5
=> x = 2.
When x = 2, y = 7. This means S = (2,7).
We already know C = (5,10).
From (5,10) to (2,7), both x and y-coordinates goes down by 3. If we go down again, we get (-1,4) which is point D.
Hence coordinates of point D = (-1,4).
Date Posted:
3 years ago
* and Points D, C and S are on the same line *
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