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3 years ago

Write an equation for a line to pass through the point (2,3) and has a slope of -4

Mathematics

1 answer:

antiseptic1488 [7]3 years ago

6 0

Answer:

y = -4x + 11

Step-by-step explanation:

Equation of a line is

y - y_1 = m ( x - x _1)

We are provided

Slope = m = -4

Points = (x _1 , y_1) = (2, 3)

Insert the values

y - y_1 = m( x - x_1)

y - 3 = -4(x - 2)

Open the bracket with -4

y - 3 = -4x + 8

y - 3 -8 = -4x

y - 11 = -4x

Following this equation of a line

y = mx + C

y - intercept point y

m - slope

x - intercept point x

C - intercept of the line

y - 11 = -4x

y = -4x + 11

Therefore, the equation of the line is

y = -4x + 11

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Read 2 more answers

what is the y-intercept of the line perpendicular to the line y=-3/4x+5 that includes the point (-3,-3)?

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Answer:

Step-by-step explanation:

For this question, your friend is the point-slope form.

Point-slope form = $y-y_1 = m(x-x_1)$

The only variables you replace are m = slope, y1 = y of a given point, and x1 = x of a given point.

First, we need to find the slope that is perpendicular to the equation $y=-\frac{3}{4}x+5$

An equation that is perpendicular to another equation have to be negative reciprocals (negative inverse) of each other.

If the slope is 3, then the negative reciprocal of 3 is $-\frac{1}{3}$ .

So, since the given slope is -3/4, the negative reciprocal is $\frac{4}{3}$ .

We will use the negative reciprocal as m since we are trying to find the equation of line that includes (-3, -3) and is perpendicular to the other equation.

We will use point slope form using the (-3, -3) points and negative reciprocal as m.

$y-y_1 = m(x-x_1)\\y-(-3) = \frac{4}{3} (x-(-3))\\y+3=\frac{4}{3}(x+3)$

Now, we need to convert this into slope-intercept form. The reason why we need to do this is because y = mx + b, where b is the y-intercept. To do this, solve for y - or in other words, isolate y..

$y+3=\frac{4}{3}(x+3)\\y+3=\frac{4x}{3}+4\\y+3-3=\frac{4x}{3}+4-3\\y=\frac{4}{3}x+1$