The point R(-3,a,-1) is the midpoint of the line segment jointing the points P(1,2,b)
and Q(c,-7,4)
Find the values of:
a=
b=
C=

Answer:

The values are:

a = -5/2

b = -6

c = -7

Step-by-step explanation:

Given

P = (x₁, y₁, z₁) = (1, 2, b)  

Q =  (x₂, y₂, z₂) = (c, -7, 4)  

As the point R(-3, a, -1) is the midpoint of the line segment jointing the points P(1,2,b)  and Q(c,-7,4), so

m = R = (x, y, z) = (-3, a, -1)

Using the mid-point formula

[tex]m\:=\:\left(\frac{x_1+x_2}{2},\:\frac{y_1+y_2}{2},\:\frac{z_1+z_2}{2}\right)[/tex]

given

(x₁, y₁, z₁) = (1, 2, b) = P

(x₂, y₂, z₂) = (c, -7, 4) = Q

m = (x, y, z) = (-3, a, -1) = R

substituting the values in the mid-point formula

[tex]m\:=\:\left(\frac{x_1+x_2}{2},\:\frac{y_1+y_2}{2},\:\frac{z_1+z_2}{2}\right)[/tex]

[tex]\left(x,\:y,\:z\right)\:=\:\left(\frac{1+c}{2},\:\frac{2+\left(-7\right)}{2},\:\frac{b+4}{2}\right)[/tex]

as (x, y, z) = (-3, a, -1), so

[tex]\left(-3,\:a,\:-1\right)\:=\:\left(\frac{1+c}{2},\:\frac{2+\left(-7\right)}{2},\:\frac{b+4}{2}\right)[/tex]

so solving 'c'

-3 = (1+c) / (2)

-3 × 2 = 1+c

1+c = -6

c = -6 - 1

c = -7

solving 'a'

a = (2+(-7)) / 2

2a = 2-7

2a = -5

a = -5/2

solving b

-1 = (b+4) / 2

-2 = b+4

b+4 = -2

b = -2-4

b = -6

Thus, the values are:

a = -5/2

b = -6

c = -7

Verification:

[tex]\left(x,\:y,\:z\right)\:=\:\left(\frac{1+c}{2},\:\frac{2+\left(-7\right)}{2},\:\frac{b+4}{2}\right)[/tex]

[tex]\left(-3,\:a,\:-1\right)\:=\:\left(\frac{1+c}{2},\:\frac{2+\left(-7\right)}{2},\:\frac{b+4}{2}\right)[/tex]

put a = -5/2, b = -6, c = -7

[tex]\left(-3,\:-\frac{5}{2},\:-1\right)\:=\:\left(\frac{1+\left(-7\right)}{2},\:-\frac{5}{2},\:\frac{\left(-6\right)+4}{2}\right)[/tex]

[tex]\left(-3,\:-\frac{5}{2},\:-1\right)\:=\:\left(\frac{-6}{2},\:-\frac{5}{2},\:\frac{-2}{2}\right)[/tex]

[tex]\left(-3,\:-\frac{5}{2},\:-1\right)\:=\:\left(-3,\:-\frac{5}{2},\:-1\right)[/tex]

to compound 50% solution in which proportion must 95% sol and 30% sol be mixed to make
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7/10
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Subtracting Fractions with Unlike Denominators
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