Question

Please help me in this​

Answer 1

Question A)

Equation of AC

Information and steps:

• ∠BAC = 90°, so AC is perpendicular to AB.
• Work out gradient of AB
• Then work out perpendicular gradient (by taking the negative reciprocal of gradient of AB)
• Substitute in perpendicular gradient, and coords for A (3, 5), into formula: y -y1 = m(x - x1)

Gradient of AB:
`(y1-y2)/(x1-x2)=(-3-5)/(-1-3)=(-8)/(-4)=2`

Perpendicular gradient = negative reciprocal of 2 =
`-(1)/(2)`

Substitute in values into formular:

`y-y1=m(x-x1)`

`y-5=-(1)/(2)(x-3)`

`2y-10=-1(x-3)`

`2y=-x+13` ← Equation of AC

Equation of BC

Information and steps:

• The gradient of BC is 1/2
• Substitute in the gradient, and the coordinates for B (-1, -3) into the formula: y - y1 = m(x -x1)

`y-y1=m(x-x1)`

`y--3=(1)/(2)(x--1)`

`y+3=(1)/(2)(x+1)`

`2y+6=x+1`

`2y=x-5` ← Equation of BC

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Question B)

Information and steps:

• Point C is the intersection of BC and AB
• Equation of BC is: 2y = x-5
• Equation of AC is: 2y = -x + 13
• Using above equations to solve a simultaneous equation - to work out the intersection of BC and AB (aka point C)

`x-5=-x+13`

`2x - 5 = 13`

`2x = 18`

`x = 9` ← The x coordinate for point C

Now substitute in value for x into the equation of either BC or AC (I'll choose AC):

`2y=x-5`

`2y=9-5`

`2y=4`

`y=2` ← The y coordinate for point C

So Coordinates of C is: (9, 2)

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Question C)

Information and steps:

• AD is perpendicular to BC
• Gradient of BC is 1/2
• Work out Perpendicular gradient of this
• Work out equation of AD, by substituting in the perpendicular gradient and the coordinates for A (3, 5) into the formula: y - y = m(x - x1)
• Notice point D is the intersection of AD and BC
• Use simultaneous equations to work out the coordinates of D

Perpendicular gradient of BC = negative reciprocal of 1/2 =
`-2`

Substitute in values into following formula:

`y-y1=m(x-x1)`

`y-5=-2(x-3)`

`y-5=-2x+6`

`y=-2x+11`

Use simultaneous equations:

Equation of BC is: 2y = -5

Equation of AD is: y = -2x+ 11 → 2y = -4x + 22 (you have to make the y's of both equations equal, so you can simultaneously solve)

`x-5 = -4x+22`

`5x-5 = 22`

`5x= 27`

`x = (27)/(5)`

`x= 5.4` ← The x coordinate of D

Now substitute in the value of x into the equation for either BC or AD (I'll choose BC again):

`2y = x-5`

`2y = 5.4-5`

`2y = 0.4`

`y=0.2` ← The y coordinate of D

So the coordinates of D is: (5.4, 0.2)

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Question D

Information and steps:

• 
  `length=\sqrt{(x1-x2)^(2)+(y1-y2)^(2)}`
• x1 = x coordinate of A (which is 3)
• x2 = x coordinate of D (which is 5.4)
• y1 = y coordinate of A (which is 5)
• y2 = y coordinate of D (which is 0.2)
• Substitute in values into formula above

Length of AD:

`length=\sqrt{(x1-x2)^(2)+(y1-y2)^(2)}`

`length=\sqrt{(3-5.4)^(2)+(5-0.2)^(2)}`

`length=\sqrt{(-2.4)^(2)+(4.8)^(2)}`

`length=√(5.76+23.04)`

`length=√(28.8)`

`length=(12√(5))/(5)`

`length = 5.367` (to 3 decimal places)

Length of AD = 5.367 (2dp)

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Answers:

A)

AC equation :
`2y=-x+13`

BC equation:
`2y=x-5`

B) Coordinates of C: (9, 2)

C) Coordinates of D: (5.4, 0.2)

D) Length of AD: 5.367 (2dp)