Straight Lines

Step 1: Find the slope of the line
The given information states: Using the identity : The slope of the line is given by:

Step 2: Use the point-slope form
The equation of a line passing through with slope is: Given point and , we get: Multiplying both sides by to clear the fraction: Rearranging:

Final Answer: The equation of the line is .

Let the equation of the straight line be:

Step 1: Find the x-intercept
The x-intercept is found by setting : Given that the x-intercept is 10: Solving for :

Step 2: Find the y-intercept
The y-intercept is found by setting :

Final Answer: The y-intercept is 5.

Step 1: Convert the given equation to slope-intercept form

Given equation:

Rearrange for :

Multiply by 2 to clear the fraction:

Step 2: Identify the slope

The slope of the given line is .

Step 3: Use point-slope form for the new line

The required line is parallel, so it has the same slope and passes through (1,3).

Using point-slope form:

Expanding:

Step 4: Convert to standard form

Thus, the correct answer is:

Given:

Point:

Slope:

Using the slope-intercept form of a line:

Substituting given values:

Thus, the y-intercept is:

4

The equation of the line is given by , where is the slope of the line.
The equation is in the form , so the slope of the given line is .
Since the two lines are perpendicular, the product of their slopes must be .
Let the slope of the line be . Then, Thus, . Now the equation of the line becomes: We are given that the line passes through the point . Substituting and into the equation :

The correct option is (D) :

The equation of a straight line is given by: where is the slope and is the y-intercept.
The line we are given, , has a slope of .
Since the required line is parallel to this line, it will have the same slope, i.e., .
Now, we are given that this line passes through the point .
We can substitute and into the equation of the line to find the value of , the y-intercept.
Substitute , , and into the equation: Simplifying: Thus, the equation of the line is:

The correct option is (A) :

We need to find the new position of the point after translating it a distance of 1 unit in the direction of the line .

Step 1: Determine the direction vector.

The line has a slope of 2, so its direction vector is . To get a unit vector in this direction, we normalize :

Step 2: Compute the translation.

Translating by 1 unit in the direction of gives:

Step 3: Verify the distance.

The distance between and is:

This confirms that the translation is correct.

Step 4: Match with the options.

The coordinates of match option (D):

The given lines are:

-----------(1)

------(2)

From the equations, we can see that the slopes of the two lines are and , respectively.

Now, let the coordinates of point P be (p, 5p + 4) (since it lies on the line y = 5x + 4) and the coordinates of point Q be (since it lies on the line

Step 2:

Find the midpoint of PQ. The midpoint of two points and is given by:

Midpoint =

Given that the midpoint of PQ is , we have:

Now, we can write two equations based on the coordinates of the midpoint:

Step 3:

Solve the system of equations to find p and q. From equation 1, we get:

From equation 2, we can simplify and

Now, let's multiply the second equation by 4 to get :

Now, we can use the first equation to solve for

Substitute this value of into the equation

Now that we have the value of p, we can find the value of q:

Step 4:

Find the equation of the line passing through P and Q. The slope of the line passing through P and Q is given by:

Using the coordinates of and ,

we get:

Now, substitute the values of p and q:

Given that

Let the plane divide the line segment joining points and in the ratio

Hence, by section formula, the coordinates of point of intersection are given by :

On the plane, the -coordinate of any point is zero.

So

,

⇒

⇒ (_Ans)

We are given the line intersecting three other lines at points and , with distances and from the point . The condition to satisfy is:

Step 1: Find the intersection points and .

1. Intersection of and :

Solve the system:

Expressing from the second equation: . Substitute into the first equation:

Then, .

Thus, .

2. Intersection of and :

Solve the system:

Expressing from the first equation: . Substitute into the second equation:

Then, .

Thus, .

3. Intersection of and :

Solve the system:

Subtract the second equation from the first:

Then, .

Thus, .

Step 2: Compute distances and from .

1. Distance to :

2. Distance to :

3. Distance to :

Step 3: Substitute into the given condition.

The condition is:

Substituting and :

Simplify each term:

Cancel common factors:

Thus, the correct value of is , corresponding to option A.

Given that

The point is equidistant from the given two lines

i.e. ; 3x−2y−5=0 and 3x+2y−5=0

Now,

Let P(h, k) be a moving point such that it is equidistant from the lines 3x−2y−5=0 then

or

or

therefore,

Hence, the path of the moving points are or ( ), which are straight lines (Ans.)

The two lines of a triangle ABC in which the line AB and AC passes through the and

Then according to the question the Equation of Bisector of the angle can be found as follows

⇒

taking Positive , the equation of bisector will be

similarly taking the negative sign , the equation will be

and as per the given option the right answer option

Given that

……………………(1)

………………………(2)

Now, multiply in equation (1)

⇒ ………………...(3)

By subtracting (1) from (3) we get ;

⇒3y-25=0

by solving this ,

putting this value in the place of in equation (1), we get;

Hence, checking the options which satisfy the value of intersection point, is 5x+4y=0

Given that:
Points are,

First find the slope of the line between the points and .

Now, the equation of the line passing through the point with slope is:

Now, check if another point lies on this line:

Now analyzing the options we can state that:

when Since we have shown that when , this statement is true. k cannot be 0 when because in that case .

First, simplify the integrand by combining the terms: Now, integrate the simplified expression: Using the integration by parts formula, let: and . Then, and .
Apply integration by parts: Let , then: Thus, the integral evaluates to:

The function returns the principal value, which must be in the range . For any , is periodic with a period of and symmetric about the y-axis.
Therefore, .
Considering , the equivalent angle in the principal range can be found by converting to a positive angle, as the cosine function is even: The angle lies within the principal range . Thus, the principal value of applied to is: