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secondary 3 | A Maths
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Pls help i do not understand this question thank you soo muchh!!
Date Posted: 3 years ago
Views: 300
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For part (i), in order to find the equation of CE, you need to find the gradient of CE and the constant.
Recall: equation of line is y =mx +c, where m is the gradient and c is a constant
Since you have the coordinates of C, you have a set of values for x and y to sub into equation of line CE
CE is perpendicular to BD. Recall: perpendicular lines that intersect are related by m1 × m2 = -1, where m1 is the gradient of one line and m2, of the other. Since they gave you the equation of BD, you have the gradient of BD (-1). That means that the gradient of CE is 1.
Sub m=1, x=6 and y=4 into y=mx+c. Then, solve for c
4 = 6 +c
c =4-6
= -2
Therefore, equation is y = x - 2
Recall: equation of line is y =mx +c, where m is the gradient and c is a constant
Since you have the coordinates of C, you have a set of values for x and y to sub into equation of line CE
CE is perpendicular to BD. Recall: perpendicular lines that intersect are related by m1 × m2 = -1, where m1 is the gradient of one line and m2, of the other. Since they gave you the equation of BD, you have the gradient of BD (-1). That means that the gradient of CE is 1.
Sub m=1, x=6 and y=4 into y=mx+c. Then, solve for c
4 = 6 +c
c =4-6
= -2
Therefore, equation is y = x - 2
done
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Date Posted:
3 years ago
For part (ii), to find coordinates if E, simply solve the two equations (for BD and CE) together to find a value of x. Sub that value of x into one of the two equations to find value for y. Those values that you found are the x-coord and y-coord of E.
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