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kiruha [24]
3 years ago
12

title=" \underline{ \underline{ \text{question}}} : " alt=" \underline{ \underline{ \text{question}}} : " align="absmiddle" class="latex-formula"> Find the equation of straight line which cuts off an intercept 2 from the Y - axis and whose perpendicular distance from the origin is 1.
​

Mathematics
2 answers:
Whitepunk [10]3 years ago
7 0

Answer:

the answer for this question is the picture hope this helps

Inga [223]3 years ago
6 0

Answer:

y=-\sqrt{3}x+2

Step-by-step explanation:

We want to find the equation of a straight line that cuts off an intercept of 2 from the y-axis, and whose perpendicular distance from the origin is 1.

We will let Point M be (x, y). As we know, Point R will be (0, 2) and Point O (the origin) will be (0, 0).

First, we can use the distance formula to determine values for M. The distance formula is given by:

\displaystyle d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Since we know that the distance between O and M is 1, d=1.

And we will let M(x, y) be (x₂, y₂) and O(0, 0) be (x₁, y₁). So:

\displaystyle 1=\sqrt{(x-0)^2+(y-0)^2}

Simplify:

1=\sqrt{x^2+y^2}

We can solve for y. Square both sides:

1=x^2+y^2

Rearranging gives:

y^2=1-x^2

Take the square root of both sides. Since M is in the first quadrant, we only need to worry about the positive case. Therefore:

y=\sqrt{1-x^2}

So, Point M is now given by (we substitute the above equation for y):

M(x,\sqrt{1-x^2})

We know that Segment OM is perpendicular to Line RM.

Therefore, their <em>slopes will be negative reciprocals</em> of each other.

So, let’s find the slope of each segment/line. We will use the slope formula given by:

\displaystyle m=\frac{y_2-y_1}{x_2-x_1}

Segment OM:

For OM, we have two points: O(0, 0) and M(x, √(1-x²)). So, the slope will be:

\displaystyle m_{OM}=\frac{\sqrt{1-x^2}-0}{x-0}=\frac{\sqrt{1-x^2}}{x}

Line RM:

For RM, we have the two points R(0, 2) and M(x, √(1-x²)). So, the slope will be:

\displaystyle m_{RM}=\frac{\sqrt{1-x^2}-2}{x-0}=\frac{\sqrt{1-x^2}-2}{x}

Since their slopes are negative reciprocals of each other, this means that:

m_{OM}=-(m_{RM})^{-1}

Substitute:

\displaystyle \frac{\sqrt{1-x^2}}{x}=-\Big(\frac{\sqrt{1-x^2}-2}{x}\Big)^{-1}

Now, we can solve for x. Simplify:

\displaystyle \frac{\sqrt{1-x^2}}{x}=\frac{x}{2-\sqrt{1-x^2}}

Cross-multiply:

x(x)=\sqrt{1-x^2}(2-\sqrt{1-x^2})

Distribute:

x^2=2\sqrt{1-x^2}-(\sqrt{1-x^2})^2

Simplify:

x^2=2\sqrt{1-x^2}-(1-x^2)

Distribute:

x^2=2\sqrt{1-x^2}-1+x^2

So:

0=2\sqrt{1-x^2}-1

Adding 1 and then dividing by 2 yields:

\displaystyle \frac{1}{2}=\sqrt{1-x^2}

Then:

\displaystyle \frac{1}{4}=1-x^2

Therefore, the value of x is:

\displaystyle \begin{aligned}\frac{1}{4}-1&=-x^2\\-\frac{3}{4}&=-x^2\\ \frac{3}{4}&=x^2\\ \frac{\sqrt{3}}{2}&=x\end{aligned}

Then, Point M will be:

\begin{aligned} \displaystyle M(x,\sqrt{1-x^2})&=M(\frac{\sqrt{3}}{2}, \sqrt{1-\Big(\frac{\sqrt{3}}{2}\Big)^2)}\\M&=(\frac{\sqrt3}{2},\frac{1}{2})\end{aligned}

Therefore, the slope of Line RM will be:

\displaystyle \begin{aligned}m_{RM}&=\frac{\frac{1}{2}-2}{\frac{\sqrt{3}}{2}-0} \\ &=\frac{\frac{-3}{2}}{\frac{\sqrt{3}}{2}}\\&=-\frac{3}{\sqrt3}\\&=-\sqrt3\end{aligned}

And since we know that R is (0, 2), R is the y-intercept of RM. Then, using the slope-intercept form:

y=mx+b

We can see that the equation of Line RM is:

y=-\sqrt{3}x+2

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