Definitions [2]
The two mutually perpendicular number lines intersecting each other at their zeroes are called rectangular axes or coordinate axes, or axes of reference.
The position of a point in a plane is expressed by a pair of numbers, one concerning the x-axis and the other concerning the y-axis. called co-ordinates.
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x → distance from y-axis (abscissa)
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y → distance from x-axis (ordinate)
Theorems and Laws [4]
Prove that the diagonals of a rectangle ABCD with vertices A(2, –1), B(5, –1), C(5, 6) and D(2, 6) are equal and bisect each other.
The vertices of the rectangle ABCD are A(2, -1), B(5, -1), C(5, 6) and D(2, 6) Now,
`"Coordinates of midpoint of" AC = ((2+5)/2 , (-1+6)/2) = (7/5 ,5/2)`
`"Coordinates of midpoint of " BD = ((5+2)/2 , (-1+6)/2)= (7/2,5/2)`
Since, the midpoints of AC and BD coincide, therefore the diagonals of rectangle ABCD bisect each other.
If the points P(x, y) is equidistant from the points A(5, 1)and B(–1, 5), prove that 3x = 2y.
As per the question, we have
AP = BP
`⇒ sqrt((x -5)^2 +(y-1)^2) = sqrt((x+1)^2 +(y-5)^2)`
`⇒(x-5)^2 +(y-1)^2 = (x+1)^2 +(y-5)^2` (Squaring both sides)
`⇒x^2 - 10x +25 + y^2 -2y +1 = x^2 +2x +1+y^2 -10y+25`
⇒ –10x – 2y = 2x – 10y
⇒ 8y = 12x
⇒ 3x = 2y
If the point (x, y) is equidistant form the points (a + b, b – a) and (a – b, a + b), prove that bx = ay.
As per the question, we have
`sqrt((x-a-b)^2 +(y-b+a)^2 ) = sqrt((x-a+b)^2 +(y-a-b)^2)`
`⇒(x-a-b)^2 +(y-b+a)^2 = (x-a+b)^2 +(y-a-b)^2` (Squaring both sides)
`⇒x^2 + (a+b)^2 -2x (a+b) +y^2 +(a-b)^2 -2y(a-b)=x^2 +(a-b)^2 -2x(a-b)+y^2 +(a+b)^2 -2y (a+b)`
`⇒-x(a+b) - y (a-b) = -x(a-b) -y(a+b)`
`⇒-xa -xb -ay +by = -xa + bx -ya-by`
⇒ by=bx
Hence, . bx = ay
Prove that the points A(–4, –1), B(–2, –4), C(4, 0) and D(2, 3) are the vertices of a rectangle.
Let A (-4,-1); B (-2,-4); C (4, 0) and D (2, 3) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a rectangle.
So we should find the lengths of opposite sides of quadrilateral ABCD.
`AB = sqrt((-2+4)^2) + (-4 + 1)^2)`
`= sqrt(4 + 9)`
`= sqrt13`
`CD = sqrt((4 - 2)^2 + (0 - 3)^2)`
`= sqrt(4 +9)`
`= sqrt13`
Opposite sides are equal. So now we will check the lengths of the diagonals.
`AC = sqrt((4 + 4)^2 + (0 + 1)^2)`
`= sqrt(64 + 1)`
`= sqrt(65)`
`BD = sqrt((2 + 2)^2 + (3 + 4)^2)`
`= sqrt(16 + 49)`
`= sqrt65`
Opposite sides are equal as well as the diagonals are equal. Hence ABCD is a rectangle.
The given points are A (-4,-1); B (-2,-4); C (4, 0) and D (2, 3) .
`AB = sqrt({-2-(-4)}^2 + { -4-(-1)}^2) = sqrt ((2)^2+(-3)^2) = sqrt(4+9) = sqrt(13) ` units
` BC = sqrt({ 4-(-2)}^2+{0-(-4)}^2) = sqrt((6)^2 +(4)^2) = sqrt(36+16) = sqrt(52) = 2 sqrt(13) units`
`CD = sqrt((2-4)^2 +(3-0)^2) = sqrt((-2)^2 +(3)^2) = sqrt(4+9) = sqrt(13) units`
`AD = sqrt({2-(-4)}^2 + {3-(-1)}^2) = sqrt((6)^2 +(4)^2) = sqrt(36+16) = sqrt(52) = 2 sqrt(13) units`
`Thus , AB = CD = sqrt(13) units and BC = AD = 2 sqrt(13) units`
Also , `AC = sqrt({4-(-4)}^2+{0-(-1)}^2) = sqrt ((8)^2+(1)^2 ) = sqrt(64+1) = sqrt(65) units`
`BD = sqrt({2-(-2)}^2 +{3-(-4)}^2) = sqrt((4)^2 +(7)^2) = sqrt(16+49) = sqrt(65) units`
Also, diagonal AC = diagonal BD
Hence, the given points form a rectanglr
Key Points
Sign Convention
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Right of y-axis → +x
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Left of y-axis → −x
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Above x-axis → +y
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Below x-axis → −y
Standard Line Results
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x = 0 → y-axis
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y = 0 → x-axis
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x = a → line parallel to the y-axis
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y = b → line parallel to the x-axis
Quadrant Reminder
| Quadrant | Sign of (x, y) |
|---|---|
| I | (+, +) |
| II | (−, +) |
| III | (−, −) |
| IV | (+, −) |
