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rodikova [14]
3 years ago
6

For the straight line defined by the points ( 3 , 51 ) (3,51) and ( 5 , 87 ) (5,87) , determine the slope ( m m ) and y-intercep

t ( b b ). Do not round the answers.
Mathematics
1 answer:
Lerok [7]3 years ago
4 0

Answer:

Intercept = - 3 ; slope = 18

Step-by-step explanation:

Given the points :

(3, 51) ; (5, 87)

x1 =3, y1 = 51, x2 = 5 ; y2 = 87

Slope intercept relation:

y = mx + c

m = slope ; c = intercept ; x and y are x, y vlayes respectively

Slope (m) = (y2 - y1) / (x2 - x1)

Slope (m) = (87 - 51) / (5 - 3)

Slope = 36 / 2

Slope (m) = 18

The intercept c :

m = 18 ; x = 5 ; y = 87

Substitute values into the slope intercept formula :

87 = 18(5) + c

87 = 90 + c

c = 87 - 90

c = - 3

Intercept = - 3

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If D is midpoint of side BC of triangle ABC. P and Q are points lying respectively on side AB and AC such that DP is parallel to
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See proof below

Step-by-step explanation:

Assume triangle ABC to have vertices at;

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The P and Q, are lying on side AB and AC, hence assume P is at (2,-4) and Q is at (4,-4) such at DP is parallel to QA

Plot the points on a graph tool and join the points to view the sketch.

To prove area of triangle CPQ is 1/4 area of ABC will be;

Find area ABC and CPQ then compare the areas.

Apply the distance formula to find the length of sides of the triangles then find the areas.

The distance formula is;

d=\sqrt{x_2-x_1)^2+(y_2-y_1)^2}

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AB=\sqrt{(2-2)^2+(-7--1)^2} =\sqrt{-6^2} =\sqrt{36} =6units

Length of side BC will be;

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1/2*4*6=12 square units

Find the lengths of all sides of triangle CPQ

Length of side PQ is half that of side BC thus PQ=2 units

Length of side PC is;

PC=\sqrt{(6-2)^2+(-7--4)^2} =\sqrt{4^2+-3^2} =\sqrt{16+9} =\sqrt{25} =5units

Length of side QC will be;

QC=\sqrt{(6-4)^2+(-7--4)^2} =\sqrt{2^2+-3^3} =\sqrt{4+9} =\sqrt{13}

QC= √13 = 3.6 units

Find area of triangle CPQ given all sides by applying the Heron's formula for area of triangle which is;

A=√s(s-a)(s-b)(s-c)  where;

A=area of the triangle

s= half the perimeter of the triangle

a=side PQ = 2 units

b=side PC = 5 units

c= side QC = 3.6 units

Finding the perimeter of triangle CPQ will be;

P=sum of all sides

P=2+5+3.6 =10.6 units

s=10.6/2 = 5.3

Area of the triangle CPQ will be;

A=\sqrt{5.3(5.3-2)(5.3-5)(5.3-3.6)} \\A=\sqrt{5.3(3.3)(0.3)(1.7)} \\A=\sqrt{8.9} =2.98

A=3.0 (1 decimal place)

Compare the areas;

Area of triangle ABC=12 square units

Area of triangle CPQ = 3 square units

Area of triangle CPQ / Area of triangle ABC = 3/12 =1/4

Thus you have proved that area of triangle CPQ is 1/4 th area of triangle ABC because 1/4 *12 =3

Learn More

Area of a triangle ;brainly.com/question/14869984

The Heron's formula : brainly.com/question/10713495

Keywords: midpoint, triangle, sides, parallel, prove , area, equal

#LearnwithBrainly

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