Mathematics, 19.12.2020 08:30 s9066115

$\underline{ \underline{ \text{question}}} :$ 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.

Find the equation of straight line which cuts off an intercept 2 from the Y - axis and whose perpe

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Answer from: vivian2020

$y=-\sqrt{3}x+2$

Step-by-step explanation:

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

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 one, 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 slopes will be negative reciprocals 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:

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

Therefore, the slope of Line RM will be:

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