Coordinate Geometry

Given: The equations of the planes are:

Step 1: Find normal vectors

Normal to the first plane:

Normal to the second plane:

Step 2: Find the cosine of the angle

The formula for the angle between two planes is:

Compute the dot product:

Compute magnitudes:

Compute :

Final Answer:

We are asked to find the area of the region bounded by the curve , the x-axis, and .

The area under the curve is given by the integral:

.

Now, compute the integral:

.

Evaluate the integral from 0 to 4:

.

The correct answer is 40.

Given equation of the line:

Rewriting in standard form:

Finding the slope of the line passing through points (-5, -2) and (4, 7):

Comparing with the equation , we get:

Substituting one point (-5, -2) into :

Calculating :

Thus, the correct answer is:

5

Step 1: General Equation of a Plane
The equation of a plane passing through a point and having normal vector is given by:

Step 2: Given Data
The point is and the normal vector is , so , , and .

Step 3: Substitute in Plane Equation
Expanding:

Final Answer: .

Step 1: Identify the standard form of the parabola
Since the axis of the parabola is parallel to the x-axis, the equation of the parabola is of the form: where is the vertex.
Given vertex: , the equation becomes:

Step 2: Find the value of
The given point lies on the parabola. Substituting , into the equation:

Step 3: Write the final equation
Substituting :

Final Answer: The equation of the parabola is .

Step 1: Find the center of the circle
The two given diameters are segments of the lines: The center of the circle is the intersection of these two lines.

Step 2: Solve for intersection point
Solving for : Substituting into : Substituting into : So, the center of the circle is .

Step 3: Write the equation of the circle
The standard equation of a circle is: Given and radius : Expanding:

Final Answer: The equation of the circle is .

Step 1: Find the center of the circle
The center of a circle whose diameter has endpoints and is given by: Given endpoints: and

Step 2: Find the radius
The radius is half the length of the diameter. The length of the diameter is: Radius is:

Step 3: Write the equation of the circle
The standard equation of a circle is: Substituting and :

Final Answer: The equation of the circle is .

Step 1: Convert the given equation into standard form
The given equation of the circle is: Divide the entire equation by 3 to simplify:

Step 2: Complete the square
Group the terms involving and :
Completing the square for :
Completing the square for :

Step 3: Write the equation in standard form
Substituting these values:
Simplifying:

Step 4: Find the radius
The standard form of a circle is: Comparing, we get:

Final Answer: The radius of the circle is .

Step 1: Find the midpoint of the given line segment

Midpoint formula:

Given points: and

Step 2: Find the slope of the given line segment

Slope formula:

Step 3: Find the slope of the perpendicular bisector

The perpendicular slope is the negative reciprocal of :

Step 4: Find the equation of the perpendicular bisector

Using point-slope form:

Multiplying everything by 2 to eliminate fractions:

Thus, the correct answer is:

We are given the equation of the curve and the equation of the normal line , which is normal to the curve at the point .
The equation of the normal line is . Rewriting this in slope-intercept form: Thus, the slope of the normal line is .
The slope of the tangent line at any point on the curve is given by the derivative of the curve : At the point , the slope of the tangent line is: Since the normal line and the tangent line are perpendicular, the product of their slopes must be : Substitute and : Simplifying this equation:
Now substitute and into the equation of the curve to find : Thus, and , so:

The correct option is (E) :

The general equation of a circle is given by: where , , and are constants. To find these constants, we use the fact that the three points , , and lie on the circle. Substituting the point into the equation: Now, the equation of the circle becomes: Substitute the point into this equation: Substitute the point into the equation: Now, we solve the system of equations: Multiplying the second equation by 3: Subtracting the first equation from this: Substitute into : Thus, the equation of the circle is: Multiplying the whole equation by 2 to eliminate the fractions:

The correct option is (E) :

The general equation of a circle is , where is the center and is the radius. We are given the equation of the circle as . The radius of the circle is given as 2, so we use the formula for the radius of a circle: where , , and . Substitute the values for and into the formula and simplify: We know that the radius is 2, so we have: Squaring both sides: Simplifying: Thus, the point lies on the circle given by the equation .

The correct option is (D) :

We are given the equation of a line and a circle:
Line: - Circle:
To find the points of intersection, we substitute the equation of the line into the equation of the circle.
Step 1: Substitute into the circle equation Substitute into the circle equation:
Step 2: Expand both terms First, expand : Now expand : Substitute both expansions into the equation: Simplify:
Step 3: Solve for Now subtract 8 from both sides: Divide by 2: Take the square root of both sides: ### Step 4: Find the corresponding values For , substitute into the line equation : For , substitute into the line equation :

The correct option is (A) :

For three lines to be concurrent, the determinant of their coefficients must be equal to 0. The given lines are: The system of equations for the lines is: We will now calculate the determinant: Expanding each of the 2x2 determinants: Substituting these values back: For the lines to be concurrent, the determinant must be 0: Solving for :

The correct option is (C) :

Let , , and be collinear. We are told that divides such that .
This condition means that divides in the ratio .
Using the section formula to find the coordinates of point , we know that: where divides in the ratio . Here and , so: Similarly, for the y-coordinate:

The correct option is (C) :

The general equation of a line whose intercepts with the coordinate axes are equal is: where is the intercept on both axes. Simplifying this, we get the equation of the line: Since the line passes through the point , we substitute these values into the equation: Thus, . Therefore, the equation of the line is:

The correct option is (B) :

The equations of the planes are given by:

These are vertical planes that are parallel to the -plane (for the first plane) and the -plane (for the second plane).

Since one plane is vertical along the -axis and the other is vertical along the -axis, the angle between these two planes is the angle between the -axis and the -axis, which is .

So, the correct option is (D) :

The equations of the two lines are given as symmetric equations. Let's first convert the symmetric form into parametric form. For the first line: This can be written as: So, the direction ratios of the first line are . For the second line: This can be written as: So, the direction ratios of the second line are . The formula for the cosine of the angle between two lines with direction ratios and is: Substitute and into the formula: Thus, .

The correct option is (D) :

Let the vertices be A(2, -6), B(5, 2), and C(-2, 2).

1. Slope of AB:

2. Slope of BC: (BC is horizontal)

3. Slope of AC:

4. Altitude from C to AB: The altitude from C is perpendicular to AB. The slope of this altitude is the negative reciprocal of : The equation of the altitude from C is .

5. Altitude from A to BC: Since BC is horizontal, the altitude from A is vertical. Its equation is simply .

6. Find the orthocenter: The orthocenter is the intersection of the altitudes. Substitute into the equation of the altitude from C:

So, the orthocenter is H(2, 1/2).

7. Line joining origin and orthocenter: The origin is O(0, 0). The line joining O and H has slope: Using the point-slope form with the origin, the equation of the line is or

Therefore, the answer is (B).

Step 1: Understand the equation.

The given equation is:

We need to determine the number of solutions for .

Step 2: Simplify and rewrite the equation.

Multiply through by 2 to eliminate the fraction:

Rearrange the equation:

Step 3: Analyze the domain.

The square root term imposes a restriction on the domain:

Thus, the equation is defined only for .

Step 4: Behavior of the two sides of the equation.

Let us analyze the two sides of the equation separately.

Left-hand side: .

• This is a cubic polynomial, which is continuous and strictly increasing for all .
• At , .
• As , .

Right-hand side: .

• This function is defined only for .
• It is continuous and strictly increasing for because the square root function is increasing.
• At , .
• As , .

Step 5: Intersection of the two functions.

Both and are continuous and strictly increasing for . Therefore, their graphs can intersect at most once in this domain.

Verify if there is an intersection:

At :

• ,
• .

Since at , and both functions increase without bound as , there must be exactly one point of intersection where .

Final Answer: 3

To find the point on the line closest to the origin, we can use the concept of projection as mentioned before. The point on the line closest to the origin will be the point where the line intersects the line passing through the origin and perpendicular to the given line.

Let's first find the direction vector of the line passing through and

Direction vector

Now, let's find the vector from the origin to any point on the line:

Vector

To find the point P on the line closest to the origin, the vector OP should be perpendicular to the direction vector of the line. Hence, their dot product should be zero:

(OP) · (Direction vector) =

We need to find the value of that satisfy this equation and lie on the line passing through .

We can use the parametric equation for the line:

Substitute these expressions into the equation

Solve for t:

Now that we have the value of t, we can find the coordinates of point P:

So, the coordinates of point P are .

Now, to find the distance squared ( ) from the origin to point P, we use the distance formula:

.

Therefore, = (_Ans).

The perpendicular bisector passes through the midpoint of the given points. Midpoint formula: Substituting and : The slope of the line joining these points is: The perpendicular bisector has the negative reciprocal slope: Equation of the perpendicular bisector: Substituting : Multiplying by 2:
Final Answer: