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Fudgin [204]
3 years ago
15

A line passes through (-7,-5) and (-5,4). Write an equation for the line in point-slope form. Rewrite the equation in standard f

orm using integers.
A. y+5= 9/2 (x-7); -9x+2y=53
B. y+7= 9/2 (x+5); -9x+2y=31
C. y+5= 9/2 (x+7); -9x+2y=53
D. y-5= 9/2 (x+7); -9x + 2y=-53
Mathematics
2 answers:
rewona [7]3 years ago
4 0
The equation:
y - y 1 =  ( y2 - y1 ) / ( x2- x1) * ( x - x1 )
y - ( - 5 ) = ( 4 + 5 ) / ( - 5 + 7 ) * ( x - ( - 7 ) )
y + 5 = 9/2 ( x + 7 )
y + 5 = 9/2 x + 63 /2  / * 2
2 y + 10 = 9 x + 63
- 9 x + 2 x = 53
Answer:
C ) y + 5 = 9/2 ( x + 7 ) ;   - 9 x + 2 y = 53
Pie3 years ago
3 0

Answer:

Option C is correct

y+5 = \frac{9}{2}(x+7)

-9x+2y=53

Step-by-step explanation:

Point slope form:

The equation of line is given by:

y-y_1=m(x-x_1)     ....[1] where m is the slope and a point (x_1, y_1) lies on the line.

Given that:

A line passes through (-7,-5) and (-5,4).

Calculate  slope:

Slope is given by:

\text{Slope} = \frac{y_2-y_1}{x_2-x_1}

Substitute the given values we have;

\text{Slope (m)} = \frac{4-(-5)}{-5-(-7)}

Simplify:

m = \frac{9}{2}

Substitute thee value of m and (-7, -5) in [1] we have;

y-(-5)=\frac{9}{2}(x-(-7))

Simplify:

y+5 = \frac{9}{2}(x+7)

⇒2(y+5) = 9(x+7)

Using distributive property :a \cdot(b+c) = a\cdot b+ a\cdot c

2y+10=9x+63

Subtract 9x from both sides we have;

-9x+2y+10=63

Subtract 10 from both sides we have;

-9x+2y=53

Therefore, an equation for the line in point-slope form is y+5 = \frac{9}{2}(x+7) and the equation in standard form using integers is -9x+2y=53

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Tomika heard that the diagonals of a rhombus are perpendicular to each other. Help her test her conjecture. Graph quadrilateral
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Answer:

a. The four sides of the quadrilateral ABCD are equal, therefore, ABCD is a rhombus

b. The equation of the diagonal line AC is y = 5 - x

The equation of the diagonal line BD is y = 5 - x

c. The diagonal lines AC and BD of the quadrilateral ABCD are perpendicular to each other

Step-by-step explanation:

The vertices of the given quadrilateral are;

A(1, 4), B(6, 6), C(4, 1) and D(-1, -1)

a. The length, l, of the sides of the given quadrilateral are given as follows;

l = \sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}

The length of side AB, with A = (1, 4) and B = (6, 6) gives;

l_{AB} = \sqrt{\left (6-4  \right )^{2}+\left (6-1  \right )^{2}} = \sqrt{29}

The length of side BC, with B = (6, 6) and C = (4, 1) gives;

l_{BC} = \sqrt{\left (1-6  \right )^{2}+\left (4-6  \right )^{2}} = \sqrt{29}

The length of side CD, with C = (4, 1) and D = (-1, -1) gives;

l_{CD} = \sqrt{\left (-1-1  \right )^{2}+\left (-1-4  \right )^{2}} = \sqrt{29}

The length of side DA, with D = (-1, -1) and A = (1,4)   gives;

l_{DA} = \sqrt{\left (4-(-1)  \right )^{2}+\left (1-(-1)  \right )^{2}} = \sqrt{29}

Therefore, each of the lengths of the sides of the quadrilateral ABCD are equal to √(29), and the quadrilateral ABCD is a rhombus

b. The diagonals are AC and BD

The slope, m, of AC is given by the formula for the slope of a straight line as follows;

Slope, \, m =\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}

Therefore;

Slope, \, m_{AC} =\dfrac{1-4}{4-1} = -1

The equation of the diagonal AC in point and slope form is given as follows;

y - 4 = -1×(x - 1)

y = -x + 1 + 4

The equation of the diagonal AC is y = 5 - x

Slope, \, m_{BD} =\dfrac{-1-6}{-1-6} = 1

The equation of the diagonal BD in point and slope form is given as follows;

y - 6 = 1×(x - 6)

y = x - 6 + 6 = x

The equation of the diagonal BD is y = x

c. Comparing the lines AC and BD with equations, y = 5 - x and y = x, which are straight line equations of the form y = m·x + c, where m = the slope and c = the x intercept, we have;

The slope m for the diagonal AC = -1 and the slope m for the diagonal BD = 1, therefore, the slopes are opposite signs

The point of intersection of the two diagonals is given as follows;

5 - x = x

∴ x = 5/2 = 2.5

y = x = 2.5

The lines intersect at (2.5, 2.5), given that the slopes, m₁ = -1 and m₂ = 1 of the diagonals lines satisfy the condition for perpendicular lines m₁ = -1/m₂, therefore, the diagonals are perpendicular.

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3 years ago
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