Learn Unit-III: Coordinate Geometry with Top Tutors

As a seasoned tutor on UrbanPro, I'm well-versed in assisting students with various math problems. UrbanPro is indeed one of the best platforms for online coaching and tuition, offering a plethora of resources and expert guidance.

Now, let's tackle your math question. You're asking for the equation of a line that is 5 units away from the origin and makes a 30-degree angle with the positive x-axis.

To solve this, we can start by recognizing that the line we're seeking is perpendicular to the line passing through the origin and making a 30-degree angle with the positive x-axis.

The slope of the line making a 30-degree angle with the x-axis is given by the tangent of 30 degrees, which is 1/√3. Therefore, the slope of the perpendicular line will be the negative reciprocal of this, which is -√3.

We know that the distance from the origin to the line is 5 units. The formula to calculate this distance for a line with equation Ax + By + C = 0 is given by |C|/√(A^2 + B^2). Since the origin is (0,0), we can simplify this to |C|/√(A^2 + B^2) = 5.

Since the line passes through the origin, its equation will be of the form y = mx. Substituting the slope (-√3) into this equation, we get y = -√3x.

Now, let's find the value of C. Since the line passes through the origin, C = 0.

Now, we can substitute the values into the distance formula: |0|/√(1^2 + (-√3)^2) = 5. Solving this equation, we find that |0|/√(1 + 3) = 5, which simplifies to 0/√4 = 5, and hence 0 = 5.

This implies that our equation y = -√3x satisfies the condition. So, the equation of the line which is at a perpendicular distance of 5 units from the origin and makes a 30-degree angle with the positive x-axis is y = -√3x.

I hope this explanation clarifies the concept for you. If you have any further questions or need additional assistance, feel free to ask!

As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is one of the best platforms for online coaching and tuition. Now, let's tackle the math problem.

To find the equation of the line perpendicular to x−7y+5=0x−7y+5=0 and passing through the point (3, 0) (since the x-intercept is given as 3), we need to first find the slope of the given line and then use the fact that perpendicular lines have slopes that are negative reciprocals.

Given the line x−7y+5=0x−7y+5=0, we can rewrite it in slope-intercept form y=mx+by=mx+b where mm is the slope and bb is the y-intercept. So, let's rearrange the given line equation:

x−7y+5=0x−7y+5=0 7y=x+57y=x+5 y=17x+57y=71x+75

So, the slope of the given line is m=17m=71.

Now, the slope of any line perpendicular to this line will be the negative reciprocal of mm. So, the slope of the perpendicular line will be −1m−m1.

mperpendicular=−117=−7mperpendicular=−711=−7

Now, we have the slope of the perpendicular line, and we know it passes through the point (3, 0). We can use the point-slope form to find the equation of the line:

y−y1=m(x−x1)y−y1=m(x−x1)

Plugging in the values, we get:

y−0=−7(x−3)y−0=−7(x−3)

Simplify:

y=−7x+21y=−7x+21

So, the equation of the line perpendicular to x−7y+5=0x−7y+5=0 and passing through the x-intercept 3 is y=−7x+21y=−7x+21.

As a seasoned tutor registered on UrbanPro, I'm well-versed in guiding students through various mathematical concepts. When it comes to solving problems like this, UrbanPro is indeed an excellent platform for students seeking online coaching and tuition.

Let's tackle this problem step by step. We have the equation of a line, y=mx+cy=mx+c, and we know that the perpendicular from the origin meets this line at the point (−1,2)(−1,2).

The equation of a line perpendicular to y=mx+cy=mx+c passing through the origin is given by y=−1mxy=−m1x. This is derived from the fact that the product of the slopes of perpendicular lines is -1.

Given that this perpendicular line meets the original line at (−1,2)(−1,2), we can equate the xx and yy coordinates to find mm.

So, when x=−1x=−1, y=2y=2, we have:

2=−1m×(−1)2=−m1×(−1) 2=1m2=m1

Solving for mm, we get m=12m=21.

Now, since the perpendicular line passes through the origin, we substitute x=0x=0 and y=0y=0 into y=−1mxy=−m1x to find the value of cc:

0=−112×0+c0=−211×0+c 0=0+c0=0+c c=0c=0

Therefore, the values of mm and cc are m=12m=21 and c=0c=0, respectively. And with UrbanPro as a resource, students can access expert guidance and support to master such mathematical concepts effectively.

∣4x0+3(0)−12∣=4

∣4x0−12∣5=45∣4x0−12∣=4

Now, we can solve for x0x0:

∣4x0−12∣=20∣4x0−12∣=20

4x0−12=±204x0−12=±20

Solving for x0x0:

When 4x0−12=204x0−12=20: 4x0=324x0=32 x0=8x0=8

When 4x0−12=−204x0−12=−20: 4x0=−84x0=−8 x0=−2x0=−2

So, the points on the x-axis that are 4 units away from the given line are x=8x=8 and x=−2x=−2.