Question:

Find the angle between the x-axis and the line joining the points (3,−1) and (4,−2).

Answer:

1. Calculate the slope of the line joining the two points, which is given by m = (y2-y1)/(x2-x1) m = (-2-(-1))/(4-3) m = -1/1

2. Calculate the angle of the line from the slope θ = tan-1(m) θ = tan-1(-1) θ = -45°

Question:

A line passes through (x1​,y1​) and (h,k). If slope of the line is m, show that k−y1​=m(h−x1​)

Answer:

Given: x1, y1, h, k, and m

Step 1: Subtract y1 from both sides of the equation. k - y1 = m(h - x1)

Step 2: Subtract k from both sides of the equation. 0 = m(h - x1) - (k - y1)

Step 3: Multiply both sides of the equation by -1. 0 = -m(h - x1) + (k - y1)

Step 4: Add m(h - x1) to both sides of the equation. m(h - x1) = k - y1

Step 5: Divide both sides of the equation by m. k - y1 = m(h - x1)

Question:

If three points (h,0),(a,b) and (0,k) lie on a line, show that ha​+kb​=1.

Answer:

1. Since the given points lie on the same line, we can write the equation of the line as y=mx+c, where m is the slope of the line and c is the y-intercept.

2. Substituting the points (h,0) and (0,k) in the equation of the line, we get:

0=mh+c and k=mc

3. Solving the above two equations, we get:

m=k/h and c=0

4. Substituting the values of m and c in the equation of the line, we get:

y=k/hx

5. Substituting the point (a,b) in the equation of the line, we get:

b=k/ha

6. Multiplying both sides by ha, we get:

ha+kb=1

Question:

Draw a quadrilateral in the Cartesian plane, whose vertices are (-4, 5), (0, 7), (5, -5) and (-4, -2). Also, find its area

Answer:

Step 1: Plot the vertices on the Cartesian plane. The vertices of the quadrilateral are (-4, 5), (0, 7), (5, -5) and (-4, -2).

Step 2: Connect the vertices in the order they are given. This will create a quadrilateral.

Step 3: Calculate the area of the quadrilateral. The formula for the area of a quadrilateral is A = 1/2(x1y2 + x2y3 + x3y4 + x4y1 - x2y1 - x3y2 - x4y3 - x1y4). Substituting the values of the vertices, the area of the quadrilateral is A = 1/2(-47 + 0-5 + 5*-2 - (-4)5 - 07 - 5*-5 - (-4)*-2) = 40.

Question:

The slope of a line is double of the slope of another line. If tangent of the angle between them is 31​, find the slopes of the lines.

Answer:

Let the slope of the first line be ’m’ and the slope of the second line be ‘2m’.

The equation of the lines in slope-intercept form is given by:

Line 1: y = mx + c Line 2: y = 2mx + c

The angle between the lines can be found using the formula:

tan θ = |m - 2m| / (1 + m2 + 2m2)

Substituting the values, we get:

tan θ = |m - 2m| / (1 + m2 + 2m2)

tan θ = |m - 2m| / 3m2

31 = |m - 2m| / 3m2

31 * 3m2 = |m - 2m|

93m2 = |m - 2m|

93m2 = m - 2m

93m2 + 2m = m

95m = m

95m = 0

m = 0

Therefore, the slope of the first line is 0 and the slope of the second line is 2 * 0 = 0.

Question:

Find the distance between P(x1​,y1​) and Q(x2​,y2​) when PQ is parallel to the y-axis.

Answer:

Step 1: Calculate the difference between the x-coordinates of the two points, x1​ and x2​.

Step 2: Subtract the smaller x-coordinate from the larger x-coordinate.

Step 3: Calculate the difference between the y-coordinates of the two points, y1​ and y2​.

Step 4: Subtract the smaller y-coordinate from the larger y-coordinate.

Step 5: The distance between P(x1​,y1​) and Q(x2​,y2​) is the absolute value of the difference between the y-coordinates.

Question:

Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P(0,−4) and B(8,0)

Answer:

1. Find the coordinates of the mid-point of the line segment joining the points P(0,−4) and B(8,0):

The mid-point of the line segment joining the points P(0,−4) and B(8,0) is (4, -2).

2. Find the slope of the line passing through the origin and the mid-point of the line segment joining the points P(0,−4) and B(8,0):

The slope of the line passing through the origin and the mid-point of the line segment joining the points P(0,−4) and B(8,0) is -1/2.

Question:

The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find the vertices of the triangle.

Answer:

1. First, determine the coordinates of the midpoint of the base. Since the midpoint of the base is at the origin, the coordinates of the midpoint are (0, 0).

2. Determine the coordinates of the other two vertices of the triangle. Since the triangle is equilateral, all sides are equal in length. Therefore, the coordinates of the other two vertices can be found by adding or subtracting 2a from the x-coordinate of the midpoint. The coordinates of the other two vertices are (2a, 0) and (-2a, 0).

3. The vertices of the triangle are (0, 0), (2a, 0), and (-2a, 0).

Question:

Without using the Pythagoras theorem, show that the points (4,4),(3,5) and (1,1) are the vertices of a right angled triangle

Answer:

Step 1: Draw a diagram of the points

Step 2: Label the points A (4,4), B (3,5) and C (1,1)

Step 3: Measure the lengths of the sides of the triangle:

AB = √((4-3)² + (4-5)²) = √(1² + (-1)²) = √2

BC = √((3-1)² + (5-1)²) = √(2² + 4²) = √20

AC = √((4-1)² + (4-1)²) = √(3² + 3²) = √18

Step 4: Since the sum of the squares of the two shorter sides is equal to the square of the longest side, the triangle is a right angled triangle.

AB² + BC² = AC²

2² + 20² = 18²

4 + 400 = 324

404 = 324

Therefore, the points (4,4), (3,5) and (1,1) are the vertices of a right angled triangle.

Question:

Without using distance formula, show that points (−2,−1),(4,0),(3,3) and (−3,2) are the vertices of a parallelogram

Answer:

Step 1: Draw a diagram of the points.

Step 2: Connect the points in order to form two pairs of opposite sides.

Step 3: Measure the lengths of the two pairs of opposite sides.

Step 4: If the lengths of the two pairs of opposite sides are equal, then the points form a parallelogram.

Question:

Find a point on x-axis which is equidistant from A(7,6) and B(−3,4) A (2,0) B (4,0) C (3,0) D (5,0)

Answer:

Answer: C (3,0)

Question:

Find the slope of the line, which makes an angle of 30o with the positive direction of y-axis measured anticlockwise.

Answer:

Step 1: Draw a line making an angle of 30o with the positive direction of the y-axis measured anticlockwise.

Step 2: Calculate the slope of the line using the formula, m = tan θ, where m is the slope and θ is the angle of the line.

Step 3: Substitute the value of θ, which is 30o, in the formula to calculate the slope.

Step 4: The slope of the line is m = tan 30o = 1/√3.

Question:

Find the value of x so that the points (x,−1),(2,1) and (4,5) are collinear.

Answer:

1. Find the equation of the line that passes through the points (x,-1), (2,1), and (4,5).

The equation of the line can be found using the slope-intercept form: y = mx + b.

The slope (m) can be found using the formula: m = (y2 - y1) / (x2 - x1).

Substituting the points (x,-1), (2,1), and (4,5):

m = (1 - (-1)) / (2 - x)

m = (2) / (2 - x)

2. Solve for x:

Rearranging the equation:

2 - x = 2 / m

2 - x = 2 / (2 / (2 - x))

2 - x = (2 - x)^2

2 - x = 4 - 4x + x^2

x^2 - 3x + 2 = 0

3. Use the quadratic formula to solve for x:

x = (3 ± √(9 - 8)) / 2

x = (3 ± 1) / 2

x = 1 or x = 2

title: “01” parent: “10 Straight Lines Exercise” draft: false

Question:

Find the angle between the x-axis and the line joining the points (3,−1) and (4,−2).

Answer:

1. First, we need to find the slope of the line joining the points (3, -1) and (4, -2). This can be done by using the formula for the slope of a line: m = (y2 - y1) ÷ (x2 - x1).

2. Plugging in the coordinates, we get m = (-2 - (-1)) ÷ (4 - 3) = -1 ÷ 1 = -1.

3. Now, we can use the formula for the angle between the x-axis and a line: tan θ = m, where θ is the angle.

4. Plugging in the slope, we get tan θ = -1.

5. Solving for θ, we get θ = tan-1(-1) = -45°.

Therefore, the angle between the x-axis and the line joining the points (3, -1) and (4, -2) is -45°.

Question:

A line passes through (x1​,y1​) and (h,k). If slope of the line is m, show that k−y1​=m(h−x1​)

Answer:

1. Rearrange the equation to make k the subject: k = m(h - x1) + y1

2. Substitute the given values: k = m(h - x1) + y1 = m(h - x1​) + y1​

3. Simplify: k = m(h - x1​) + y1​ = mh - mx1​ + y1​ = mh - mx1​ + y1​ = m(h - x1​) + y1​

Question:

If three points (h,0),(a,b) and (0,k) lie on a line, show that ha​+kb​=1.

Answer:

1. Using the point-slope formula, we can write the equation of the line passing through the three points as: y - 0 = (k - 0)/(0 - h) * (x - h)

2. Simplifying, we have: y = -(k/h)x + k

3. Substituting the coordinates of the given points in the equation of the line, we obtain: 0 - 0 = -(k/h)a + k b - 0 = -(k/h)h + k

4. Simplifying, we get: k = ha b = ha + k

5. Multiplying the two equations, we get: ha * (ha + k) = ha * k + k^2

6. Simplifying, we obtain: ha + kb = 1

Question:

Draw a quadrilateral in the Cartesian plane, whose vertices are (-4, 5), (0, 7), (5, -5) and (-4, -2). Also, find its area

Answer:

1. Draw the quadrilateral in the Cartesian plane, connecting the given vertices in the given order.

2. Calculate the lengths of the four sides of the quadrilateral by using the distance formula.

3. Calculate the area of the quadrilateral using the formula A = 1/2(x1y2 + x2y3 + x3y4 + x4y1 - x2y1 - x3y2 - x4y3 - x1y4).

4. Substitute the values of the coordinates of the vertices in the formula and calculate the area.

Question:

The slope of a line is double of the slope of another line. If tangent of the angle between them is 31​, find the slopes of the lines.

Answer:

Let the slope of the first line be m1 and the slope of the second line be m2.

Since the slope of the first line is double of the slope of the second line, we have m1 = 2m2.

Now, let θ be the angle between the two lines.

We know that, tanθ = m1 - m2/1 + m1m2

Substituting m1 = 2m2, we get

tanθ = 2m2 - m2/1 + 2m2^2

=> tanθ = m2 (2 - 1 + 2m2)/1 + 2m2^2

=> tanθ = m2 (3 + 2m2)/1 + 2m2^2

Given that tanθ = 31, we have

31 = m2 (3 + 2m2)/1 + 2m2^2

=> 31 (1 + 2m2^2) = m2 (3 + 2m2)

=> 31 + 62m2^2 = 3m2 + 6m2^2

=> 59m2^2 = 31

=> m2^2 = 31/59

=> m2 = ± √(31/59)

=> m2 = ± 0.7745966692414834

Therefore, the slope of the second line is m2 = ± 0.7745966692414834

Hence, the slope of the first line is m1 = 2m2 = ± 1.549193338482967

Question:

Find the distance between P(x1​,y1​) and Q(x2​,y2​) when PQ is parallel to the y-axis.

Answer:

Step 1: Calculate the difference between the x-coordinates of the two points, x1​-x2​.

Step 2: Set the difference equal to zero, as the line is parallel to the y-axis.

Step 3: Calculate the difference between the y-coordinates of the two points, y1​-y2​.

Step 4: The distance between P and Q is equal to the absolute value of the difference between the y-coordinates, |y1​-y2​|.

Question:

Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P(0,−4) and B(8,0)

Answer:

Step 1: Find the coordinates of the mid-point of the line segment joining the points P(0,−4) and B(8,0).

The mid-point of the line segment joining the points P(0,−4) and B(8,0) is (4, -2).

Step 2: Find the slope of the line passing through the origin and the mid-point of the line segment.

The slope of the line passing through the origin and the mid-point of the line segment is -1/2.

Question:

The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find the vertices of the triangle.

Answer:

Step 1: The vertices of an equilateral triangle are three points that are all equidistant from each other.

Step 2: The mid-point of the base is at the origin (0,0).

Step 3: The side of the triangle is 2a, so the distance from the origin to each vertex is a.

Step 4: The vertices of the triangle are (a, 0), (-a, 0), and (0, a√3).

Question:

Without using the Pythagoras theorem, show that the points (4,4),(3,5) and (1,1) are the vertices of a right angled triangle

Answer:

Step 1: Draw a triangle with the given points as the vertices.

Step 2: Calculate the length of the sides of the triangle by using the distance formula.

Step 3: The length of the sides of the triangle will be:

Side AB = √((4-3)² + (4-5)²) = √2

Side BC = √((3-1)² + (5-1)²) = √8

Side AC = √((4-1)² + (4-1)²) = √9

Step 4: Check if the triangle is a right angled triangle by using the Pythagorean theorem.

Step 5: The Pythagorean theorem states that the square of the length of the longest side is equal to the sum of the squares of the other two sides.

So, 9 = 2 + 8

Therefore, the points (4,4), (3,5) and (1,1) form a right angled triangle.

Question:

Without using distance formula, show that points (−2,−1),(4,0),(3,3) and (−3,2) are the vertices of a parallelogram

Answer:

1. Draw the four points on a coordinate plane
2. Connect the points in order to form a quadrilateral
3. Measure the lengths of the sides of the quadrilateral
4. Determine if opposite sides are parallel by measuring the slopes of the sides
5. If the slopes of the opposite sides are equal, then the four points form a parallelogram

Question:

Find a point on x-axis which is equidistant from A(7,6) and B(−3,4) A (2,0) B (4,0) C (3,0) D (5,0)

Answer:

Answer: C (3,0)

Question:

Find the slope of the line, which makes an angle of 30o with the positive direction of y-axis measured anticlockwise.

Answer:

Step 1: Draw a line making an angle of 30o with the positive direction of y-axis measured anticlockwise.

Step 2: Label the line with two points (A and B) and draw a right angle triangle with the line and the positive direction of y-axis.

Step 3: Find the length of the opposite side (AB) and adjacent side (y-axis).

Step 4: Calculate the slope of the line using the formula m = AB/y-axis.

Question:

Find the value of x so that the points (x,−1),(2,1) and (4,5) are collinear.

Answer:

Step 1: Calculate the slopes of the lines between the points.

Slope between (x, -1) and (2, 1): m1 = (1 - (-1)) / (2 - x)

Slope between (2, 1) and (4, 5): m2 = (5 - 1) / (4 - 2)

Step 2: Set the slopes equal to each other and solve for x.

m1 = m2 (1 - (-1)) / (2 - x) = (5 - 1) / (4 - 2) (2 - x) (4 - 2) = (5 - 1) 2x - 4 = 4 2x = 8 x = 4