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Rom4ik [11]
2 years ago
11

8.) A line passes through the points (k, 4) and (3, -6). The line is perpendicular to

Mathematics
1 answer:
amm18122 years ago
3 0

Answer:

K = 43

Step-by-step explanation:

We'll begin by determining the gradient of the equation 5y + 4x = 8. This can be obtained as follow:

5y + 4x = 8

Rearrange

5y = 8 – 4x

5y = –4x + 8

Comparing 5y = –4x + 8 with y = mx + c, the gradient m is –4

Next, we shall determine the gradient of the line perpendicular to the line with equation 5y = 8 – 4x.

This can be obtained as follow:

For perpendicular lines, their gradient is given by:

m1 × m2 = – 1

With the above formula, we can obtain the gradient of the line as follow:

m1 × m2 = – 1

m1 = –4

–4 × m2 = – 1

Divide both side by –4

m2 = –1/–4

m2 = 1/4

Finally, we shall determine the value of k as follow:

Coordinate => (k, 4) and (3, –6)

x1 coordinate = k

y1 coordinate = 4

x2 coordinate = 3

y2 coordinate = –6

Gradient (m) = 1/4

m = (y2 – y1) / (x2 – x1)

1/4 = (–6 – 4) / (3 – K)

1/4 = –10 /(3 – K)

Cross multiply

3 – K = 4 × –10

3 – K = –40

Collect like terms

– K = – 40 –3

–k = –43

Divide both side by – 1

K = –43/–1

k = 43

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Answer:

AB = \sqrt{a^2 + b^2-2abCos\ C}

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

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Make h the subject:

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a^2 = h^2 + x^2

Make x^2 the subject

x^2 = a^2 - h^2

Substitute aSin\ C for h

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Factorize

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In trigonometry:

Cos^2C = 1-Sin^2C

So, we have that:

x^2 = a^2 Cos^2\ C

Take square roots of both sides

x= aCos\ C

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AB^2 = h^2 + (b-x)^2

Open bracket

AB^2 = h^2 + b^2-2bx+x^2

Substitute x= aCos\ C and h = aSin\ C in AB^2 = h^2 + b^2-2bx+x^2

AB^2 = h^2 + b^2-2bx+x^2

AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2

Open Bracket

AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C

Reorder

AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C

Factorize:

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Sin^2C + Cos^2 = 1

So, we have that:

AB^2 = a^2 * 1 + b^2-2abCos\ C

AB^2 = a^2 + b^2-2abCos\ C

Take square roots of both sides

AB = \sqrt{a^2 + b^2-2abCos\ C}

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