Question:

Find the equation of a straight line: with slope −2 and intersecting the x−axis at a distance of 3 units to the left of origin.

Answer:

Step 1: The equation of a straight line is y = mx + c, where m is the slope and c is the y-intercept.

Step 2: In this case, the slope is -2.

Step 3: The line intersects the x-axis at a distance of 3 units to the left of the origin. This means that the y-intercept is -3.

Step 4: Substituting the values of m and c in the equation of a straight line, we get: y = -2x - 3

Question:

Find the equation of the line which intersects the y-axis at a distance of 2 units above the origin and makes an angle of 30o with the positive direction of the x-axis.

Answer:

1. First, identify the slope of the line. The slope of the line is m = tan(30°) = √3/3.

2. Then, use the point-slope form of the equation of a line, which is: y - y1 = m(x - x1)

3. Substitute the values for m and (x1, y1), which are m = √3/3 and (x1, y1) = (0, 2): y - 2 = √3/3(x - 0)

4. Simplify the equation: y = 2 + √3/3x

Question:

Find equation of the line passing through the point (−4,3) with slope 1/2

Answer:

Step 1: Find the slope of the line using the given point and the given slope. Slope = m = (y2 - y1)/(x2 - x1) = (3 - 0)/(-4 - 0) = -3/4

Step 2: Write the equation of the line in slope-intercept form. y = mx + b

Step 3: Substitute the slope and the given point into the equation. y = (-3/4)x + b 3 = (-3/4)(-4) + b

Step 4: Solve for b. b = 15/4

Step 5: Substitute the value of b into the equation. y = (-3/4)x + (15/4)

Answer: The equation of the line passing through the point (−4,3) with slope 1/2 is y = (-3/4)x + (15/4).

Question:

Find equation of the line passing through (0,0) with slope m.

Answer:

Answer: The equation of the line passing through (0,0) with slope m can be written as y = mx.

Question:

P(a,b) is the mid-point of a line segment between axes. Show that equation of the line is x/a+y​/b=2

Answer:

Step 1: Let the line passing through the point P(a,b) be represented by the equation y = mx + c

Step 2: Since P(a,b) is the mid-point of the line segment, the coordinates of the other end point must be (0,0).

Step 3: Substituting the coordinates of the points P(a,b) and (0,0) in the equation of the line, we get

b = ma

Step 4: Dividing both sides of the equation by ab, we get

1/a + m/b = 2

Step 5: Rearranging the equation, we get

m/b = 2 - 1/a

Step 6: Multiplying both sides of the equation by ab, we get

m = ab(2 - 1/a)

Step 7: Substituting the value of m in the equation of the line, we get

y = ab(2 - 1/a)x + c

Step 8: Since the point P(a,b) lies on the line, substituting the coordinates of the point in the equation of the line, we get

b = ab(2 - 1/a)a + c

Step 9: Simplifying the above equation, we get

b = 2ab - c

Step 10: Substituting the value of b in the equation of the line, we get

y = ab(2 - 1/a)x + (2ab - b)

Step 11: Simplifying the above equation, we get

y = ab(2 - 1/a)x + ab

Step 12: Dividing both sides of the equation by ab, we get

y/b = (2 - 1/a)x + 1

Step 13: Multiplying both sides of the equation by b, we get

y = b(2 - 1/a)x + b

Step 14: Rearranging the equation, we get

x/a + y/b = 2

Hence, the equation of the line is x/a + y/b = 2.

Question:

By using the concept of equation of a line, prove that the three points (3,0), (-2, -2) and (8, 2) are collinear.

Answer:

1. Start by writing the equation of a line in the form y = mx + b.

2. Substitute the x and y values of the first point (3, 0) into the equation to solve for b.

3. Substitute the x and y values of the second point (-2, -2) into the equation to solve for m.

4. Substitute the x and y values of the third point (8, 2) into the equation to check if the equation is true.

5. If the equation is true, then the three points are collinear.

Question:

The perpendicular from the origin to a line meets it at the point (2,9), find the equation of the line.

Answer:

1. First, we need to find the slope of the line. To do this, we can use the slope formula: m = (y2 - y1)/(x2 - x1)

Substituting the given point (2,9): m = (9 - 0)/(2 - 0) m = 9/2

2. Now, we can use the point-slope form of the line equation to find the equation of the line: y - y1 = m(x - x1)

Substituting the given point (2,9) and the slope (9/2): y - 9 = (9/2)(x - 2)

3. Finally, we can simplify the equation to get the final answer: y = 9/2x - 9/2

Question:

The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L=124.942 when C=20 and L=125.134 when C=110, express L in terms of C.

Answer:

L = mC + b

124.942 = m(20) + b

125.134 = m(110) + b

Subtract b from both sides:

124.942 - b = m(20) 125.134 - b = m(110)

Subtract m(20) from both sides:

125.134 - 124.942 = m(110) - m(20)

0.192 = 90m

Divide both sides by 90:

m = 0.002133

Substitute m into the equation 124.942 = m(20) + b:

124.942 = 0.002133(20) + b

Subtract 0.002133(20) from both sides:

124.942 - 0.042660 = b

b = 124.89954

L = 0.002133C + 124.89954

Question:

Passing through (2,2√3) and inclined with the x-axis at an angle of 75∘

Answer:

Step 1: Draw a line passing through (2,2√3).

Step 2: Measure the angle of inclination between the line and the x-axis.

Step 3: Adjust the line so that it is inclined to the x-axis at an angle of 75∘.

Question:

The perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive x−axis is 30o.Find the equation of the line?

Answer:

Answer: Step 1: Find the x-intercept and y-intercept of the line.

x-intercept = 0

y-intercept = 5

Step 2: Use the slope-intercept form of the equation of a line to find the equation.

Slope = tan 30o = √3/3

Equation of the line = y = (√3/3)x + 5

Question:

The vertices of △PQR are P(2,1),Q(−2,3) and R(4,5). Find equation of the median through the vertex R.

Answer:

Step 1: Find the coordinates of the midpoint of the line segment PQ.

Let the midpoint of the line segment PQ be M(x, y).

We can use the midpoint formula to find the coordinates of M:

x = (2 + (−2))/2 = 0 y = (1 + 3)/2 = 2

Therefore, the coordinates of M are (0, 2).

Step 2: Find the coordinates of the midpoint of the line segment QR.

Let the midpoint of the line segment QR be N(x, y).

We can use the midpoint formula to find the coordinates of N:

x = (4 + (−2))/2 = 1 y = (5 + 3)/2 = 4

Therefore, the coordinates of N are (1, 4).

Step 3: Find the equation of the line passing through the midpoints M and N.

We can use the two-point form of the equation of a line to find the equation of the line passing through M and N:

y − 2 = (4 − 2)/(1 − 0) (x − 0)

y − 2 = 2(x − 0)

y = 2x + 2

Therefore, the equation of the line passing through M and N is y = 2x + 2.

Step 4: Find the equation of the median through the vertex R.

Since the coordinates of the vertex R are (4, 5), we can substitute these coordinates into the equation of the line passing through M and N to find the equation of the median through the vertex R:

5 = 2(4) + 2

5 = 8 + 2

5 = 10

Therefore, the equation of the median through the vertex R is y = 10.

Question:

Find the equation of the line passing through (−3,5) and perpendicular to the line through the points (2,5) and (−3,6).

Answer:

1. Find the slope of the line through (2,5) and (−3,6). m = (6-5)/(-3-2) = -1

2. The slope of the line perpendicular to this line is the negative reciprocal of the slope. m = -1/1 = -1

3. Use the point-slope form of the equation of a line: y - y1 = m(x - x1) y - 5 = -1(x + 3)

4. Simplify the equation: y - 5 = -x - 3

5. Solve for y: y = -x - 3 + 5

6. The equation of the line passing through (−3,5) and perpendicular to the line through the points (2,5) and (−3,6) is y = -x - 3 + 5.

Question:

A line perpendicular to the line segment joining the points (1,0) and (2,3) divides it in the ratio 1:n. Find the equation of the line.

Answer:

Step 1: Find the slope of the line segment joining the two points.

Slope = (y2 - y1) / (x2 - x1)

Slope = (3 - 0) / (2 - 1)

Slope = 3/1

Step 2: Find the slope of the line perpendicular to the line segment.

Slope of perpendicular line = -1 / slope of original line

Slope of perpendicular line = -1 / 3

Step 3: Find the equation of the line in slope-intercept form.

y = mx + b

y = (-1/3)x + b

Step 4: Use the point (1,0) to find the value of b.

0 = (-1/3) * 1 + b

b = 1

Step 5: Substitute the value of b into the equation.

y = (-1/3)x + 1

Therefore, the equation of the line is y = (-1/3)x + 1.

Question:

Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2,3).

Answer:

Answer: Step 1: The equation of a line passing through the point (2,3) is y = mx + c, where m is the slope of the line and c is the y-intercept.

Step 2: To find the slope of the line, calculate the rise (change in y) over the run (change in x) between the two points. The rise is 3 - 0 = 3 and the run is 2 - 0 = 2, so the slope is 3/2.

Step 3: Substitute the slope (3/2) into the equation of the line: y = (3/2)x + c.

Step 4: To find the y-intercept, substitute the coordinates of the point (2,3) into the equation and solve for c: 3 = (3/2)2 + c → c = -3/2.

Step 5: Substitute the slope (3/2) and the y-intercept (-3/2) into the equation of the line: y = (3/2)x - 3/2.

Therefore, the equation of the line that cuts off equal intercepts on the coordinate axes and passes through the point (2,3) is y = (3/2)x - 3/2.

Question:

Find equation of the line passing through the point (2,2) and cutting off intercepts on the axes whose sum is 9.

Answer:

Answer:

Step 1: Let the equation of the line be y = mx + c

Step 2: Substitute the given point (2,2) in the equation to find the value of c

2 = m*2 + c

c = 2 - 2m

Step 3: Let the intercepts on the x-axis and y-axis be x1 and y1 respectively.

Step 4: Since the intercepts are on the axes, their respective coordinates are (x1, 0) and (0, y1).

Step 5: Substitute the coordinates in the equation of the line to find the values of m and c.

0 = m*x1 + c

2 - 2m = m*0 + c

Step 6: Solve the above two equations to find the values of m and c.

m = -2 and c = 2

Step 7: Substitute the values of m and c in the equation of the line to get the required equation.

y = -2x + 2

Step 8: Use the equation to find the intercepts.

x1 = 9/2 and y1 = -4

Step 9: Verify that the sum of the intercepts is 9.

x1 + y1 = 9/2 - 4 = 9

Question:

The owner of a milk store finds that he can sell 980 liters of milk each week at Rs. 14 per lit. and 1220 liters of milk each week at Rs. 16 per lit. Assuming a liner relationship between selling price and demand , how many liters could you sell weakly at Rs. 17 per liter?

Answer:

Step 1: Calculate the slope of the line.

Slope = (1220 - 980)/(16 - 14) = 120/2 = 60

Step 2: Calculate the y-intercept of the line.

y-intercept = 980 - (14 * 60) = 980 - 840 = 140

Step 3: Substitute the given value of x (17) in the equation of the line.

y = 60x + 140

y = 60(17) + 140

y = 1020

Step 4: The number of liters of milk that can be sold at Rs. 17 per liter is 1020 liters.

Question:

Write the equations for the x-and y-axes.

Answer:

The equation for the x-axis is y = 0

The equation for the y-axis is x = 0

Question:

Find equation of line passing through the points (−1,1) and (2,−4).

Answer:

Step 1: Recall that the equation of a line in slope-intercept form is y = mx + b, where m is the slope and b is the y-intercept.

Step 2: Calculate the slope (m) of the line. m = (y2 - y1)/(x2 - x1) m = (-4 - 1)/(2 - (-1)) m = -5/3

Step 3: Substitute the slope (m) and one of the given points into the equation of a line to calculate the y-intercept (b). b = y - mx b = 1 - (-5/3)(-1) b = 1 + 5/3 b = 8/3

Step 4: Substitute the slope (m) and the y-intercept (b) into the equation of a line to get the equation of the line passing through the given points. y = -5/3x + 8/3

Question:

Find equation of the line through the point (0,2) making an angle 2π​/3 with the positive x-axis. Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin.

Answer:

Equation of the line through the point (0,2) making an angle 2π/3 with the positive x-axis: y - 2 = tan(2π/3)(x - 0) => y = tan(2π/3)x + 2

Equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin: y - (-2) = tan(2π/3)(x - 0) => y = tan(2π/3)x - 2

Question:

Point R (h, k) divides a line segment between the axes in the ratio 1 : 2. Find equation of the line. A 2kx+ hy = 3hk B 2k<+ hy 2hk C 2kx - hy = 3hk D None of the above

Answer:

Answer: C 2kx - hy = 3hk

title: “02” parent: “10 Straight Lines Exercise” draft: false

Question:

Find the equation of a straight line: with slope −2 and intersecting the x−axis at a distance of 3 units to the left of origin.

Answer:

Step 1: The equation of a straight line is y = mx + c, where m is the slope and c is the y-intercept.

Step 2: In this case, the slope is −2.

Step 3: The y-intercept is 3 units to the left of the origin, so c = -3.

Step 4: Substituting the values of m and c in the equation of a straight line, we get: y = -2x - 3

Question:

Find the equation of the line which intersects the y-axis at a distance of 2 units above the origin and makes an angle of 30o with the positive direction of the x-axis.

Answer:

Answer: Step 1: Identify the coordinates of the point of intersection of the line with the y-axis.

Since the point lies 2 units above the origin, the coordinates of the point of intersection are (0,2).

Step 2: Find the slope of the line.

The angle made by the line with the positive direction of the x-axis is 30o. Therefore, the slope of the line is tan30o = 1/√3.

Step 3: Find the equation of the line using the point-slope form.

The equation of the line in the point-slope form is y - 2 = (1/√3) (x - 0).

Step 4: Simplify the equation.

The equation can be simplified to y - 2 = (1/√3)x.

Question:

Find equation of the line passing through the point (−4,3) with slope 1/2

Answer:

Step 1: The equation of a line in slope-intercept form is given by y = mx + b, where m is the slope and b is the y-intercept.

Step 2: Substitute the given values for m and the coordinates of the given point in the equation.

Step 3: y = (1/2)x + b

Step 4: Substitute the x-coordinate of the given point in the equation.

Step 5: 3 = (1/2)(-4) + b

Step 6: Solve for b.

Step 7: b = 7/2

Therefore, the equation of the line passing through the point (−4,3) with slope 1/2 is y = (1/2)x + (7/2).

Question:

Find equation of the line passing through (0,0) with slope m.

Answer:

Answer: The equation of the line passing through (0,0) with slope m is y = mx.

Question:

P(a,b) is the mid-point of a line segment between axes. Show that equation of the line is x/a+y​/b=2

Answer:

Answer: Step 1: Let P(a,b) be the mid-point of the line segment between two axes Step 2: Let P1(x1,y1) and P2(x2,y2) be the two end points of the line segment Step 3: Since P(a,b) is the mid-point, then the coordinates of P1 and P2 can be determined by: P1(a/2,b/2) and P2(3a/2,3b/2) Step 4: The equation of the line can be determined by using the point-slope form of a line equation: y-y1=m(x-x1) Step 5: Substituting the coordinates of P1 and P2 in the equation, we get: y-b/2=m(x-a/2) Step 6: Now, we can calculate the slope of the line using the coordinates of P1 and P2: m=(3b/2-b/2)/(3a/2-a/2) Step 7: Substituting the value of m in the equation, we get: y-b/2=((3b/2-b/2)/(3a/2-a/2))(x-a/2) Step 8: Simplifying the equation, we get: y/b=(x/a)(3/2-1/2) Step 9: Simplifying further, we get: y/b=x/a Step 10: Finally, we can write the equation of the line as: x/a+y/b=2

Question:

By using the concept of equation of a line, prove that the three points (3,0), (-2, -2) and (8, 2) are collinear.

Answer:

Step 1: Draw a line connecting the three points (3,0), (-2, -2) and (8, 2).

Step 2: Calculate the slope of the line by using the formula m = (y2 - y1) / (x2 - x1).

Step 3: Substitute the coordinates of the three points in the formula to calculate the slope of the line.

m = (2 - 0) / (8 - 3) m = 2/5

Step 4: Calculate the equation of the line by using the formula y = mx + c.

Step 5: Substitute the slope (m) and the coordinates of one of the points in the formula to calculate the equation of the line.

y = (2/5)x + c y = (2/5)3 + c c = -2

Step 6: Substitute the value of c in the equation of the line to get the final equation.

y = (2/5)x - 2

Step 7: Since the equation of the line is the same for all the three points, it can be concluded that the three points (3,0), (-2, -2) and (8, 2) are collinear.

Question:

The perpendicular from the origin to a line meets it at the point (2,9), find the equation of the line.

Answer:

1. First, calculate the slope of the line using the point (2,9). Slope = m = (y2-y1)/(x2-x1) = (9-0)/(2-0) = 9/2

2. Then, calculate the y-intercept of the line using the point (2,9). y-intercept = b = y - mx = 9 - (9/2)2 = 9 - 4.5 = 4.5

3. Finally, use the slope and y-intercept to find the equation of the line. Equation of the line: y = (9/2)x + 4.5

Question:

The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L=124.942 when C=20 and L=125.134 when C=110, express L in terms of C.

Answer:

L = mC + b

Substitute the two given points into the equation:

124.942 = m(20) + b

125.134 = m(110) + b

Solve for m and b:

m = (125.134 - 124.942) / (110 - 20)

m = 0.092

b = 124.942 - 0.092(20)

b = 124.742

Therefore,

L = 0.092C + 124.742

Question:

Passing through (2,2√3) and inclined with the x-axis at an angle of 75∘

Answer:

Step 1: Draw an x-axis and a y-axis on a graph paper.

Step 2: Mark the point (2,2√3) on the graph paper.

Step 3: Draw a line from the point (2,2√3) that is inclined with the x-axis at an angle of 75∘.

Question:

The perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive x−axis is 30o.Find the equation of the line?

Answer:

1. Let (x,y) be the coordinates of any point on the line.

2. Since the perpendicular distance from the origin is 5 units,

5² = x² + y²

3. Also, the angle made by the perpendicular with the positive x-axis is 30o

tan 30o = y/x

4. Solving the above two equations, we get

x = 5/√3

y = 5/2

5. Therefore, the equation of the line is

y = (5/√3)x + 5/2

Question:

The vertices of △PQR are P(2,1),Q(−2,3) and R(4,5). Find equation of the median through the vertex R.

Answer:

1. Find the midpoint of the line segment PQ. The midpoint of PQ can be found by taking the average of the x and y coordinates of P and Q.

Midpoint of PQ = ( (2 + (-2))/2, (1 + 3)/2) Midpoint of PQ = (0, 2)

2. Find the equation of the line through the midpoint and the vertex R. The equation of the line through the midpoint (0,2) and the vertex R(4,5) can be found using the point-slope form of the equation of a line.

Slope = (5 - 2)/(4 - 0) = 3/4

Equation of the line through the midpoint and the vertex R = y - 2 = (3/4)(x - 0)

3. The equation of the median through the vertex R is the equation of the line found in step 2.

Equation of the median through the vertex R = y - 2 = (3/4)(x - 0)

Question:

Find the equation of the line passing through (−3,5) and perpendicular to the line through the points (2,5) and (−3,6).

Answer:

Step 1: Find the slope of the line through (2, 5) and (−3, 6).

Slope = (6 - 5) / (−3 - 2) = -1/5

Step 2: Find the slope of the line perpendicular to the line through (2, 5) and (−3, 6).

Slope of perpendicular line = -5/1

Step 3: Find the equation of the line passing through (−3, 5) and with a slope of -5/1.

Equation of the line: y - 5 = -5/1 (x + 3)

y = -5/1x - 8

Question:

A line perpendicular to the line segment joining the points (1,0) and (2,3) divides it in the ratio 1:n. Find the equation of the line.

Answer:

Step 1: Find the slope of the line segment joining the two points.

Slope = (3-0)/(2-1) = 3

Step 2: The slope of the perpendicular line is the negative reciprocal of the slope of the line segment, which is -1/3.

Step 3: The equation of the line is y = -1/3x + b.

Step 4: Substitute the coordinates of one of the points in the equation.

(1,0): 0 = -1/3(1) + b

Step 5: Solve for b.

b = 1/3

Step 6: The equation of the line is y = -1/3x + 1/3.

Question:

Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2,3).

Answer:

Step 1: The equation of a line in the form of y = mx + c is needed.

Step 2: The slope of the line (m) can be calculated using the point-slope form of the equation of a line.

Step 3: Substitute the values of x and y into the point-slope form to calculate the slope (m).

Step 4: Substitute the calculated slope (m) and the coordinates of the given point (2,3) into the point-slope form to calculate the value of c.

Step 5: Substitute the values of m and c into the equation of the line y = mx + c to get the required equation.

Answer: The equation of the line is y = -1/2x + 5.

Question:

Find equation of the line passing through the point (2,2) and cutting off intercepts on the axes whose sum is 9.

Answer:

Answer: Step 1: We need to find the slope of the line passing through the given point (2,2).

Slope = m = (y2 - y1)/(x2 - x1)

m = (2 - 2)/(2 - 2)

m = 0

Step 2: The equation of the line passing through the point (2,2) is given by y = 2.

Step 3: The equation of the line cutting off intercepts on the axes whose sum is 9 is given by x + y = 9.

Step 4: Substituting the value of y from Step 2 in Step 3, we get x + 2 = 9

Step 5: Solving the equation, we get x = 7.

Hence, the equation of the line passing through the point (2,2) and cutting off intercepts on the axes whose sum is 9 is x = 7 and y = 2.

Question:

The owner of a milk store finds that he can sell 980 liters of milk each week at Rs. 14 per lit. and 1220 liters of milk each week at Rs. 16 per lit. Assuming a liner relationship between selling price and demand , how many liters could you sell weakly at Rs. 17 per liter?

Answer:

1. Find the equation of the linear relationship between selling price and demand:

Demand = (980 - 1220)/(14 - 16) * (Selling Price - 16) + 1220

2. Substitute the given selling price of Rs. 17 into the equation to calculate the demand:

Demand = (980 - 1220)/(14 - 16) * (17 - 16) + 1220

Demand = 1360 liters

Question:

Write the equations for the x-and y-axes.

Answer:

The equation for the x-axis is: x = 0

The equation for the y-axis is: y = 0

Question:

Find equation of line passing through the points (−1,1) and (2,−4).

Answer:

1. First, identify the slope of the line.

Slope = m = (y2-y1)/(x2-x1)

m = (-4 - 1)/(2 - (-1))

m = -5/3

2. Next, use the point-slope formula to find the equation of the line.

y - y1 = m(x - x1)

y - 1 = -5/3(x - (-1))

y - 1 = -5/3x + 5/3

3. Finally, rearrange the equation to solve for y.

y = -5/3x + 5/3 + 1

y = -5/3x + 8/3

Question:

Find equation of the line through the point (0,2) making an angle 2π​/3 with the positive x-axis. Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin.

Answer:

Equation of line through (0,2) making an angle 2π/3 with the positive x-axis:

y - 2 = tan(2π/3)(x - 0) y - 2 = (√3/3)(x - 0) y = (√3/3)x + 2

Equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin:

y - (-2) = tan(2π/3)(x - 0) y - (-2) = (√3/3)(x - 0) y = (√3/3)x - 2

Question:

Point R (h, k) divides a line segment between the axes in the ratio 1 : 2. Find equation of the line. A 2kx+ hy = 3hk B 2k<+ hy 2hk C 2kx - hy = 3hk D None of the above

Answer:

Answer: C. 2kx - hy = 3hk