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larisa86 [58]
3 years ago
7

1. Given points A(3, -5) and B(19, -1), find the coordinates of point C that sit 3/8 of the way along line AB, closer to A than

to B.
2. Given points A(3, -5) and B(19, -1), find the coordinates of point C such that CB/AC = 1/7

Mathematics
1 answer:
zaharov [31]3 years ago
8 0

1. C(x, y) = (7.3, –3.9)

2. C(x, y) = (17, –1.5)

Solution:

Question 1:

Let the points are A(3, –5) and B(19, –1).

C is the point that on the segment AB in the fraction \frac{3}{8}.

Point divides segment in the ratio formula:

$C(x, y)=\left(\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n}\right)

Here, x_1=3, y_1=-5, x_2=19, y_2=-1 and m = 3, n = 8

$C(x, y)=\left(\frac{3\times19+8\times3}{3+8} , \frac{3\times(-1)+8\times(-5)}{3+8}\right)

           $=\left(\frac{57+24}{11} , \frac{-3-40}{11}\right)

           $=\left(\frac{81}{11} , \frac{-43}{11}\right)

C(x, y) = (7.3, –3.9)

Question 2:

Let the points are A(3, –5) and B(19, –1).

C is the point that on the segment AB in the fraction \frac{3}{8}.

Point divides segment in the ratio formula:

$C(x, y)=\left(\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n}\right)

Here, x_1=3, y_1=-5, x_2=19, y_2=-1 and m = 7, n = 1

$C(x, y)=\left(\frac{7\times19+1\times3}{7+1} , \frac{7\times(-1)+1\times(-5)}{7+1}\right)

           $=\left(\frac{133+3}{8} , \frac{-7-5}{8}\right)

           $=\left(\frac{136}{8} , \frac{-12}{8}\right)

C(x, y) = (17, –1.5)

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See Below.

Step-by-step explanation:

Statements:                                                           Reasons:

\displaystyle 1)\text{ } \Delta APB \text{ and } \Delta AQC \text{ are equilateral triangles}      Given

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3)\text{ } m \angle QAC = 60                                                     Definition of equilateral.

4)\text{ } m\angle PAB = m\angle QAC                                          Substitution

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7)\text{ } m\angle QAB=m\angle PAB+m\angle BAC                       Substitution

\displaystyle 8)\text{ } m\angle PAC=m\angle QAB                                         Substitution

9)\text{ } PA=BA                                                          Definition of equilateral

10)\text{ } AC=AQ                                                        Definition of equilateral

\displaystyle 11)\text{ } \Delta PAC \cong \Delta BAQ                                            Side-Angle-Side Congruence*

\displaystyle 12)\text{ } PC=BQ                                                        CPCTC

* SAS Congruence:

PA = BA

∠PAC = ∠QAB

AC = AQ

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