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secondary 3 | A Maths
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I need help with part ii
Date Posted: 3 years ago
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This is very similar to finding the midpoint of a line segment.
https://brilliant.org/wiki/section-formula/
To quote,
If point P (x,y) lies on line segment AB (between points A and B) and satisfies AP : PB = m : n, then we say that P divides AB internally in the ratio m : n.
The point of division has the coordinates P( (mx₂ + nx₁)/(m + n) , (my₂ + ny₁)/(m + n) ) , whereby :
A has coordinates (x₁,y₁) and B has coordinates (x₂, y₂)
So for this question, P(-2,-4) and Q(5,10). PR : RQ = 2 : 5
Coordinates of R are :
( (2(5) + 5(-2))/(2 + 5) , (2(10) + 5(-4) )
= (0/7 , 0/7)
= (0,0)
https://brilliant.org/wiki/section-formula/
To quote,
If point P (x,y) lies on line segment AB (between points A and B) and satisfies AP : PB = m : n, then we say that P divides AB internally in the ratio m : n.
The point of division has the coordinates P( (mx₂ + nx₁)/(m + n) , (my₂ + ny₁)/(m + n) ) , whereby :
A has coordinates (x₁,y₁) and B has coordinates (x₂, y₂)
So for this question, P(-2,-4) and Q(5,10). PR : RQ = 2 : 5
Coordinates of R are :
( (2(5) + 5(-2))/(2 + 5) , (2(10) + 5(-4) )
= (0/7 , 0/7)
= (0,0)
Alternative method (longer way)
Let the coordinates of R be (x,y)
Length of PR = √((-2 - x)² + (-4 - y)²)
= √(4 + 4x + x² + 16 + 8y + y²)
= √(x² + y² + 4x + 8y + 20)
Length of RQ = √((5 - x)² + (10 - y)²)
= √(25 - 10x + x² + 100 - 20y + y²)
= √(x² + y² - 10x - 20y + 125)
PQ = √((5 - (-2))² + (10 - (-4))²)
= √(49 + 196)
= √245
Since PR : RQ = 2 : 5,
Then PR : PQ = 2 : (2+5) = 2 : 7 , which means that PR = 2/7 PQ
And RQ : PQ = 5 : (2 + 5) = 5 : 7, which means that QR = 5/7 PQ
So,
√(x² + y² + 4x + 8y + 20) = 2/7 √245
Squaring both sides,
x² + y² + 4x + 8y + 20 = 4/49 (245) = 20
x² + y² + 4x + 8y = 0 ①
And,
√(x² + y² - 10x - 20y + 125) = 5/7 √245
Squaring both sides,
x² + y² - 10x - 20y + 125 = 25/49 (245) = 125
x² + y² - 10x - 20y = 0 ②
① = ②
x² + y² + 4x + 8y = x² + y² - 10x - 20y
4x + 8y = -10x - 20y
28y = -14x
y = -½ x
Sub y = -½x into ①,
x² + (-½x)² + 4x + 8(-½x) = 0
x² + ¼x² + 4x - 4x = 0
5/4 x² = 0
x² = 0
x = 0
Then y = -½(0) = 0
∴ R(0,0)
Let the coordinates of R be (x,y)
Length of PR = √((-2 - x)² + (-4 - y)²)
= √(4 + 4x + x² + 16 + 8y + y²)
= √(x² + y² + 4x + 8y + 20)
Length of RQ = √((5 - x)² + (10 - y)²)
= √(25 - 10x + x² + 100 - 20y + y²)
= √(x² + y² - 10x - 20y + 125)
PQ = √((5 - (-2))² + (10 - (-4))²)
= √(49 + 196)
= √245
Since PR : RQ = 2 : 5,
Then PR : PQ = 2 : (2+5) = 2 : 7 , which means that PR = 2/7 PQ
And RQ : PQ = 5 : (2 + 5) = 5 : 7, which means that QR = 5/7 PQ
So,
√(x² + y² + 4x + 8y + 20) = 2/7 √245
Squaring both sides,
x² + y² + 4x + 8y + 20 = 4/49 (245) = 20
x² + y² + 4x + 8y = 0 ①
And,
√(x² + y² - 10x - 20y + 125) = 5/7 √245
Squaring both sides,
x² + y² - 10x - 20y + 125 = 25/49 (245) = 125
x² + y² - 10x - 20y = 0 ②
① = ②
x² + y² + 4x + 8y = x² + y² - 10x - 20y
4x + 8y = -10x - 20y
28y = -14x
y = -½ x
Sub y = -½x into ①,
x² + (-½x)² + 4x + 8(-½x) = 0
x² + ¼x² + 4x - 4x = 0
5/4 x² = 0
x² = 0
x = 0
Then y = -½(0) = 0
∴ R(0,0)
Second alternative method : similar triangles
We can take PQ to be the hypotenuse of a right-angled triangle.
Since R lies on PQ, PR and RQ are also the hypotenuses of right-angled triangles which are similar to the bigger one.
(The gradient is the same for PQ,PR and RQ which means the ratios/proportions of the rise and run (i.e their horizontal and vertical side) are the same)
Horizontal distance from x-coordinate of P to that of Q = 5 - (-2) = 7 unit's
Vertical distance from y-coordinate of P to that of Q = 10 - (-4) = 14 units
Since PR = 2/7 of PQ ,
Horizontal distance from x-coordinate of P to that of R = 7 units x 2/7 = 2 units
Vertical distance from y-coordinate of P to that of R = 14 units x 2/7 = 4 units
Therefore,
x-coordinate of R = -2 + 2 = 0
y-coordinate of R = -4 + 4 = 0
∴R(0,0)
OR
Since RQ = 5/7 of PQ ,
Horizontal distance from x-coordinate of Q to that of R = 7 units x 5/7 = 5 units
Vertical distance from y-coordinate of Q to that of R = 14 units x 5/7 = 10 units
Therefore,
x-coordinate of R = 5 - 5 = 0
y-coordinate of R = 10 - 10 = 0
∴R(0,0)
We can take PQ to be the hypotenuse of a right-angled triangle.
Since R lies on PQ, PR and RQ are also the hypotenuses of right-angled triangles which are similar to the bigger one.
(The gradient is the same for PQ,PR and RQ which means the ratios/proportions of the rise and run (i.e their horizontal and vertical side) are the same)
Horizontal distance from x-coordinate of P to that of Q = 5 - (-2) = 7 unit's
Vertical distance from y-coordinate of P to that of Q = 10 - (-4) = 14 units
Since PR = 2/7 of PQ ,
Horizontal distance from x-coordinate of P to that of R = 7 units x 2/7 = 2 units
Vertical distance from y-coordinate of P to that of R = 14 units x 2/7 = 4 units
Therefore,
x-coordinate of R = -2 + 2 = 0
y-coordinate of R = -4 + 4 = 0
∴R(0,0)
OR
Since RQ = 5/7 of PQ ,
Horizontal distance from x-coordinate of Q to that of R = 7 units x 5/7 = 5 units
Vertical distance from y-coordinate of Q to that of R = 14 units x 5/7 = 10 units
Therefore,
x-coordinate of R = 5 - 5 = 0
y-coordinate of R = 10 - 10 = 0
∴R(0,0)
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