Write an equation for the line that is parallel to y = 2x + 6 and passes through (-1,5).

Mathematics

1 answer:

Fantom [35]3 years ago

6 0

Answer:

y=2x+7

Step-by-step explanation:

A linear equation is typically organized in the form y=mx+b, where m is the slope of the line and b is the y-intercept (the y-coordinate of the point on the line that crosses the y-axis).

Parallel lines always have the same slope. When we examine the given equation, y=2x+6, we can identify immediately that 2 is in the place of m, the slope. This means that the slope of the line we're trying to solve for is also 2. So far, our equation looks like this:

y=2x+b

The last thing we need to calculate is b, the y-intercept. We can do this by plugging in the given point (-1,5) as x and y.

y=2x+b

5=2(-1)+b

5=-2+b

Add 2 to both sides to isolate b

5+2=-2+b+2

7=b

Now that we know b, we plug that into our original equation and now we have a final equation of:

y=2x+7

I hope this helps!

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Given:

diamong [38]

Answer:

10-5 $\sqrt{2}$

Step-by-step explanation:

As per the attached figure, right angled $\triangle MDL$ has an inscribed circle whose center is $I$ .

We have joined the incenter $I$ to the vertices of the $\triangle MDL$ .

Sides MD and DL are equal because we are given that $\angle M = \angle L = 45 ^\circ.$

Formula for area of a $\triangle = \dfrac{1}{2} \times base \times height$

As per the figure attached, we are given that side a = 10.

Using pythagoras theorem, we can easily calculate that side ML = 10 $\sqrt{2}$

Points P,Q and R are at $90 ^\circ$ on the sides ML, MD and DL respectively so IQ, IR and IP are heights of $\triangle$ MIL, $\triangle$ MID and $\triangle$ DIL.

Also,

$\text {Area of } \triangle MDL = \text {Area of } \triangle MIL +\text {Area of } \triangle MID+ \text {Area of } \triangle DIL$

$\dfrac{1}{2} \times 10 \times 10 = \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10\sqrt2\\\Rightarrow r = \dfrac {10}{2+\sqrt2} \\\Rightarrow r = \dfrac{5\sqrt2}{\sqrt2+1}\\\text{Multiplying and divinding by }(\sqrt2 +1)\\\Rightarrow r = 10-5\sqrt2$