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Gnom [1K]
3 years ago
6

1. Write the standard form of the line that passes through the given points. (7, -3) and (4, -8)

Mathematics
1 answer:
Sveta_85 [38]3 years ago
5 0

Answer:

1. -5x+3y+44=0

2. 2x+y-2=0

3. 2x+y-4=0

Step-by-step explanation:

Standard form of a line is Ax+By+C=0.

If a line passing through two points then the equation of line is

y-y_1=m(x-x_1)

where, m is slope, i.e.,m=\dfrac{y_2-y_1}{x_2-x_1}.

1.

The line passes through the points (7,-3) and (4,-8). So, the equation of line is

y-(-3)=\dfrac{-8-(-3)}{4-7}(x-7)

y+3=\dfrac{-5}{-3}(x-7)

y+3=\dfrac{5}{3}(x-7)

3(y+3)=5(x-7)

3y+9=5x-35

-5x+3y+9+35=0

-5x+3y+44=0

Therefore, the required equation is -5x+3y+44=0.

2.

We need to find the equation of the line, in standard form, that has a y-intercept of 2 and is parallel to 2 x + y =-5.

Slope of the line : m=\dfrac{-\text{Coefficient of x}}{\text{Coefficient of y}}=\dfrac{-2}{1}=-2

Slope of parallel lines are equal. So, the slope of required line is -2 and it passes through the point (0,2).

Equation of line is

y-2=-2(x-0)

y-2=-2x

2x+y-2=0

Therefore, the required equation is 2x+y-2=0.

3.

We need to find the equation of the line, in standard form, that has an x-intercept of 2 and is parallel to 2x + y =-5.

From part 2, the slope of this line is -2. So, slope of required line is -2 and it passes through the point (2,0).

Equation of line is

y-0=-2(x-2)

y=-2x+4

2x+y-4=0

Therefore, the required equation is 2x+y-4=0.

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