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secondary 3 | A Maths
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How to do question iv?
Date Posted: 7 years ago
Views: 419
(i) Given the two points C & M, plot a line passing these two points. This line will have an X-axis intercept.
The gradient of the line drawn is the ratio opposite/adjacent of any right-angle triangle drawn from the line to the positive X-axis. From there, obtain your angle with the trigonometry function.
(ii) Refer to the gradient of the line CM. If the line CM is perpendicular to AB, then gradient of AB * gradient of CM = -1. Otherwise its not perpendicular.
(iii) M is midpoint of CP implies that if we draw C-M as a vector and map it twice to the coordinate of C, we will get the coordinate of P.
In short, (5,5) - (2,4) = (3,1) : Vector C-M. Then M-P also has the same value (3,1). Let the coordinates of P be (x1,y1). By the above, (2,4) - (x1,y1) = (3,1).
(iv) Split the quadrilateral into two separate triangles: ABC and ABQ. Plot a rough point Q in the diagram to 'see' these two triangles.
Calculate the Area of the triangle ABC to determine the Area of ABQ. To do this, take the length AB as the base and length CM as the height of triangle ABC. Area of ABC + Area of ABQ = 50 units square.
Let the coordinates of Q be (x2,y2).
Area of ABQ = length AB * length QM. Call this simultaneous equation 1.
Q also lies on the line drawn in (i); that is to say it satisfy the equation that was formed in part (i) of the form y = mx + c, where you subtitute in (x1,y1) as the coordinates. Call this simultaneous equation 2.
You are now given two simultaneous equation, solve for x1 and y1. Coordinate of Q is found.
...
Overall the question is tedious, but do-able. you just need to take your time to construct the equations.
The gradient of the line drawn is the ratio opposite/adjacent of any right-angle triangle drawn from the line to the positive X-axis. From there, obtain your angle with the trigonometry function.
(ii) Refer to the gradient of the line CM. If the line CM is perpendicular to AB, then gradient of AB * gradient of CM = -1. Otherwise its not perpendicular.
(iii) M is midpoint of CP implies that if we draw C-M as a vector and map it twice to the coordinate of C, we will get the coordinate of P.
In short, (5,5) - (2,4) = (3,1) : Vector C-M. Then M-P also has the same value (3,1). Let the coordinates of P be (x1,y1). By the above, (2,4) - (x1,y1) = (3,1).
(iv) Split the quadrilateral into two separate triangles: ABC and ABQ. Plot a rough point Q in the diagram to 'see' these two triangles.
Calculate the Area of the triangle ABC to determine the Area of ABQ. To do this, take the length AB as the base and length CM as the height of triangle ABC. Area of ABC + Area of ABQ = 50 units square.
Let the coordinates of Q be (x2,y2).
Area of ABQ = length AB * length QM. Call this simultaneous equation 1.
Q also lies on the line drawn in (i); that is to say it satisfy the equation that was formed in part (i) of the form y = mx + c, where you subtitute in (x1,y1) as the coordinates. Call this simultaneous equation 2.
You are now given two simultaneous equation, solve for x1 and y1. Coordinate of Q is found.
...
Overall the question is tedious, but do-able. you just need to take your time to construct the equations.
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