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: $\frac{y1-y2}{x1-x2}=\frac{-3-5}{-1-3}=\frac{-8}{-4}=2$

Perpendicular gradient = negative reciprocal of 2 = $-\frac{1}{2}$

Substitute in values into formular:

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

$y-5=-\frac{1}{2}(x-3)$

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

$2y=-x+13$   ←  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)

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

$y--3=\frac{1}{2}(x--1)$

$y+3=\frac{1}{2}(x+1)$

$2y+6=x+1$

$2y=x-5$  ← 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)

$x-5 = -4x+22$

$5x-5 = 22$

$5x= 27$

$x = \frac{27}{5}$

$x= 5.4$    ← 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:

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

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

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

$length=\sqrt{5.76+23.04}$

$length=\sqrt{28.8}$

$length=\frac{12\sqrt{5}}{5}$

$length = 5.367$  (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)