Question
Determine the equation of the line that passes through 
Hint:
We are given two points and we need to find the equation of the line passing through them. The equation of a line passing through two points (a, b) and (c, d) is

The correct answer is: 8x - y + 18 = 0
Step by step solution:
Let the given points be denoted by
(a, b) = (-3, -6)
(c, d) = (2, 34)
The equation of a line passing through two points and is

Using the above points, we have

Simplifying the above equation, we have


Cross multiplying, we get
5(y - 34) = 40(x - 2)
Expanding the factors, we have
5y - 170 = 40 x -80
Taking all the terms in the left hand side, we have
-40x + 5y - 170 + 80 = 0
Finally, the equation of the line is
-40x + 5y - 90=0
Dividing the equation throughout by(- 5), we get
This is the required equation.
.
The equation of a line passing through two points and is
Using the above points, we have
Simplifying the above equation, we have
Cross multiplying, we get
Expanding the factors, we have
Taking all the terms in the left hand side, we have
Finally, the equation of the line is
Dividing the equation throughout by(- 5), we get
This is the required equation.
.
We can simplify the equation in any other way and we would still reach the same equation. The general form of an equation in two variables is given by a + by + c=0, where a, b, c are real numbers. The student is advised to remember all the different forms of a line, like, slope-intercept form, axis-intercept form, etc
Related Questions to study
Is the given conjecture correct? Provide arguments to support your answer.
The product of any two consecutive even numbers is 1 less than a perfect square
Finding the product of 2n and (2n+2)
2n(2n+2) = 2n(2n) + 2n(2)
2n(2n+2) = 4n2 + 4n
The given conjecture is “The product of any two consecutive even numbers is 1 less than a perfect square”
So, add 1 to 4n2 + 4n
4n2 + 4n +1 = (2n)2 + 2(2n)(1) + 12
= (2n + 1)2
So, we can see that the perfect square of 2n+1 is 4n2 + 4n +1 and 1 less than 4n2 + 4n +1 is a product of any two consecutive even numbers.
Final Answer:
Hence, the given conjecture is right and we have proved it by Deductive reasoning.
Is the given conjecture correct? Provide arguments to support your answer.
The product of any two consecutive even numbers is 1 less than a perfect square
Finding the product of 2n and (2n+2)
2n(2n+2) = 2n(2n) + 2n(2)
2n(2n+2) = 4n2 + 4n
The given conjecture is “The product of any two consecutive even numbers is 1 less than a perfect square”
So, add 1 to 4n2 + 4n
4n2 + 4n +1 = (2n)2 + 2(2n)(1) + 12
= (2n + 1)2
So, we can see that the perfect square of 2n+1 is 4n2 + 4n +1 and 1 less than 4n2 + 4n +1 is a product of any two consecutive even numbers.
Final Answer:
Hence, the given conjecture is right and we have proved it by Deductive reasoning.
Identify a linear polynomial.
We have to identify the linear polynomial from the given four options in the question
Step 1 of 1:
A linear polynomial is defined as any polynomial expressed in the form of an equation of
From the option We can see that option B is in the form of
So, Option B is correct.
Identify a linear polynomial.
We have to identify the linear polynomial from the given four options in the question
Step 1 of 1:
A linear polynomial is defined as any polynomial expressed in the form of an equation of
From the option We can see that option B is in the form of
So, Option B is correct.
Find the value of x. Identify the theorem used to find the answer.

- Hint:
- Mid-point theorem:
- According to mid-point theorem, in triangle, the line segment which joins the midpoint of two sides is parallel to third side and is equal to half of third side,
- Step by step explanation:
- Given:
BM = MC,
hence M is midpoint of BC.
BN = NA,
hence N is midpoint of AB.
MN = x
AC = 24
- Step 1:
The line segment MN joins midpoint of AB and BC
So,
According to midpoint theorem,
MN is parallel to AC and
MN =
MN =
MN = 12
∴ x = 12.
Final Answer:
x = 12.
Find the value of x. Identify the theorem used to find the answer.

- Hint:
- Mid-point theorem:
- According to mid-point theorem, in triangle, the line segment which joins the midpoint of two sides is parallel to third side and is equal to half of third side,
- Step by step explanation:
- Given:
BM = MC,
hence M is midpoint of BC.
BN = NA,
hence N is midpoint of AB.
MN = x
AC = 24
- Step 1:
The line segment MN joins midpoint of AB and BC
So,
According to midpoint theorem,
MN is parallel to AC and
MN =
MN =
MN = 12
∴ x = 12.
Final Answer:
x = 12.
identify a cubic polynomial.
- We have to identify the cubic polynomial from the given four options.
We know that a cubic polynomial is a polynomial with degree 3.
Option A:
x + 3
The degree of this polynomial 1.
Option B:
The degree of the polynomial is 4.
Option C:
The degree of this polynomial is 3.
Option D:
The degree of this polynomial 2.
identify a cubic polynomial.
- We have to identify the cubic polynomial from the given four options.
We know that a cubic polynomial is a polynomial with degree 3.
Option A:
x + 3
The degree of this polynomial 1.
Option B:
The degree of the polynomial is 4.
Option C:
The degree of this polynomial is 3.
Option D:
The degree of this polynomial 2.
Find the value of 𝑚 & 𝑛 to make a true statement.
(𝑚𝑥 + 𝑛𝑦)2 = 4𝑥2 + 12𝑥𝑦 + 9𝑦2
(mx + ny)(mx + ny) = mx(mx + ny) + ny(mx + ny)
= mx(mx) + mx(ny) + ny(mx) + ny(ny)
= m2x2 + mnxy + mnxy + n2y2
= m2x2 + 2mnxy + n2y2
Now, m2x2 + 2mnxy + n2y2 = 4𝑥2 + 12𝑥𝑦 + 9𝑦2
Comparing both sides, we get
m2 = 4, n = 9, 2mn = 12
So, m = +2 or -2 , n = +3 or -3
Considering 2mn = 12, there are two combinations possible
- m = +2 and n = +2
- m = -2 and n = -2
Hence, the values of (m, n) are (2,2) and (-2,-2).
Find the value of 𝑚 & 𝑛 to make a true statement.
(𝑚𝑥 + 𝑛𝑦)2 = 4𝑥2 + 12𝑥𝑦 + 9𝑦2
(mx + ny)(mx + ny) = mx(mx + ny) + ny(mx + ny)
= mx(mx) + mx(ny) + ny(mx) + ny(ny)
= m2x2 + mnxy + mnxy + n2y2
= m2x2 + 2mnxy + n2y2
Now, m2x2 + 2mnxy + n2y2 = 4𝑥2 + 12𝑥𝑦 + 9𝑦2
Comparing both sides, we get
m2 = 4, n = 9, 2mn = 12
So, m = +2 or -2 , n = +3 or -3
Considering 2mn = 12, there are two combinations possible
- m = +2 and n = +2
- m = -2 and n = -2
Hence, the values of (m, n) are (2,2) and (-2,-2).
37. In an academic contest correct answers earn 12 points and incorrect answers lose 5
points. In the final round, school A starts with 165 points and gives the same number
of correct and incorrect answers. School B starts with 65 points and gives no incorrect answers and the same number of correct answers as school A. The game ends with the two schools tied.
i)Which equation models the scoring in the final round and the outcome of the contest
- Step by step explanation:
correct answer = 12 points
incorrect answers = -5 points
School A starts with 165 points and gives the same number of correct and incorrect answers.
School B starts with 65 points and gives no incorrect answers and the same number of correct answers as school A
○ Step 1:
○ Let the number of correct answers given by school A be x.
So, the number of incorrect answers is also x.
At school A starts with 165 points. After giving x correct and a incorrect answers points will be
School B starts with 65 points. Schools are given the same number of correct answers as school A and no incorrect answers. So, their points will be
65 + 12x
○ Step 2:
○ As both schools tied
∴ 165 + 12x - 5x = 65 + 12x
- Final Answer:
Option B. 165 + 12x - 5x = 65 + 12x
37. In an academic contest correct answers earn 12 points and incorrect answers lose 5
points. In the final round, school A starts with 165 points and gives the same number
of correct and incorrect answers. School B starts with 65 points and gives no incorrect answers and the same number of correct answers as school A. The game ends with the two schools tied.
i)Which equation models the scoring in the final round and the outcome of the contest
- Step by step explanation:
correct answer = 12 points
incorrect answers = -5 points
School A starts with 165 points and gives the same number of correct and incorrect answers.
School B starts with 65 points and gives no incorrect answers and the same number of correct answers as school A
○ Step 1:
○ Let the number of correct answers given by school A be x.
So, the number of incorrect answers is also x.
At school A starts with 165 points. After giving x correct and a incorrect answers points will be
School B starts with 65 points. Schools are given the same number of correct answers as school A and no incorrect answers. So, their points will be
65 + 12x
○ Step 2:
○ As both schools tied
∴ 165 + 12x - 5x = 65 + 12x
- Final Answer:
Option B. 165 + 12x - 5x = 65 + 12x
Find the value of x. Identify the theorem used to find the answer.

- Hints:
- Perpendicular bisector theorem
- According to perpendicular bisector theorem, in the triangle, any point on perpendicular bisector is at equal distance from both end points of the line segment on which it is drawn.
- Step by step explanation:
- Given:
AC = 4x – 4
AD is perpendicular bisector at BC.
- Step 1:
- In
A is point on AD
A is equidistant from B and C.
So,
AB = AC
2x = 4x – 4
4 = 4x – 2x
4 = 2x
x = 2
- Final Answer:
Perpendicular bisector theorem is used.
Find the value of x. Identify the theorem used to find the answer.

- Hints:
- Perpendicular bisector theorem
- According to perpendicular bisector theorem, in the triangle, any point on perpendicular bisector is at equal distance from both end points of the line segment on which it is drawn.
- Step by step explanation:
- Given:
AC = 4x – 4
AD is perpendicular bisector at BC.
- Step 1:
- In
A is point on AD
A is equidistant from B and C.
So,
AB = AC
2x = 4x – 4
4 = 4x – 2x
4 = 2x
x = 2
- Final Answer:
Perpendicular bisector theorem is used.
Where is the circumcentre located in any right triangle? Write a coordinate proof of this result.
- Hints:
- Distance between two points having coordinates (x1, y1) and (x2, y2) is given by formula:
- Distance =
- Step by step explanation:
- Step 1:
- Let triangle ABO,
O = (0, 0)
A = (2a, 0)
B = (0, 2b).
- Step 1:
- Let triangle ABO, where:
The midpoint of BC is given by,
So, the perpendicular bisector will intersect BC at M (a, b).
Equation of line BC is
And point M (a, b) satisfy above equation
b = b
Hence, point M (a, b) lies on BC.
- Final Answer:
Where is the circumcentre located in any right triangle? Write a coordinate proof of this result.
- Hints:
- Distance between two points having coordinates (x1, y1) and (x2, y2) is given by formula:
- Distance =
- Step by step explanation:
- Step 1:
- Let triangle ABO,
O = (0, 0)
A = (2a, 0)
B = (0, 2b).
- Step 1:
- Let triangle ABO, where:
The midpoint of BC is given by,
So, the perpendicular bisector will intersect BC at M (a, b).
Equation of line BC is
And point M (a, b) satisfy above equation
b = b
Hence, point M (a, b) lies on BC.
- Final Answer:
Ayush is choosing between two health clubs. Health club 1: Membership R s 22 and
Monthly fee R s 24.50. Health club 2: Membership R s 47.00 , monthly fee R s 18.25
. After how many months will the total cost for each health club be the same ?
- Hint:
- Step by step explanation:
Health club 1: Membership R s 22 and Monthly fee R s 24.50
Health club 2: Membership R s 47.00 , monthly fee R s 18.25.
○ Step 1:
○ Let the number of months after which the total cost is equal be x.
So, for Health club 1:
After x months, total cost will be =
Health club 2:
After x months, total cost will be =
○ Step 2:
○ Equalize both costs to get number of months
- Final Answer:
Ayush is choosing between two health clubs. Health club 1: Membership R s 22 and
Monthly fee R s 24.50. Health club 2: Membership R s 47.00 , monthly fee R s 18.25
. After how many months will the total cost for each health club be the same ?
- Hint:
- Step by step explanation:
Health club 1: Membership R s 22 and Monthly fee R s 24.50
Health club 2: Membership R s 47.00 , monthly fee R s 18.25.
○ Step 1:
○ Let the number of months after which the total cost is equal be x.
So, for Health club 1:
After x months, total cost will be =
Health club 2:
After x months, total cost will be =
○ Step 2:
○ Equalize both costs to get number of months
- Final Answer:
Find the gradient and y- Intercept of the line 
Gradient is also called the slope of the line. The slope intercept form of the equation of the line is y = mx + c, where m is the slope of the line and c is the y-intercept. First we convert the given equation in this form. Further, compare the equation with the standard form to get the slope and the y-intercept.
Step by step solution:
The given equation of the line is
x + 2y = 14
We need to convert this equation in the slope-intercept form of the line, which is
y = mx + c
Rewriting the given equation, we have
2y = 14 - x
Dividing by 2, we get
Simplifying, we get
Comparing the above equation with , we get
Thus, we get
Gradient =
y-intercept = 7
Note:
We can find the slope and y-intercept directly from the general form of the equation too; slope =
Find the gradient and y- Intercept of the line 
Gradient is also called the slope of the line. The slope intercept form of the equation of the line is y = mx + c, where m is the slope of the line and c is the y-intercept. First we convert the given equation in this form. Further, compare the equation with the standard form to get the slope and the y-intercept.
Step by step solution:
The given equation of the line is
x + 2y = 14
We need to convert this equation in the slope-intercept form of the line, which is
y = mx + c
Rewriting the given equation, we have
2y = 14 - x
Dividing by 2, we get
Simplifying, we get
Comparing the above equation with , we get
Thus, we get
Gradient =
y-intercept = 7
Note:
We can find the slope and y-intercept directly from the general form of the equation too; slope =
We can express any constant in the variable form without changing its value as
- We have given a constant
- We have to find how we can express any constant 𝑘 in the variable form without changing its value.
We have given a constant k
We have to find how we can express any constant 𝑘 in the variable form without changing its value
We know that x0 = 1
So After multiplying it with any constant, it will not change its value.
So,kx0 is the answer
Hence, Option C is correct.
We can express any constant in the variable form without changing its value as
- We have given a constant
- We have to find how we can express any constant 𝑘 in the variable form without changing its value.
We have given a constant k
We have to find how we can express any constant 𝑘 in the variable form without changing its value
We know that x0 = 1
So After multiplying it with any constant, it will not change its value.
So,kx0 is the answer
Hence, Option C is correct.
x0 = ?
x0 = ?
State and prove the Perpendicular Bisector Theorem.
- To prove:
- Perpendicular Bisector Theorem:
- Proof:
- Statement:
- According to perpendicular bisector theorem, in the triangle, any point on perpendicular bisector is at equal distance from both end points of the line segment on which it is drawn.
- Step 1:
Let arbitrary point C on perpendicular bisector.
In which,
CD is perpendicular bisector on AB.
Hence,
AD = DB
CD = CD (common)
So, according to SAS rule
Hence,
CA = CB
So, any point on perpendicular bisector is at equal distance from end points of line segment.
Hence proved.
State and prove the Perpendicular Bisector Theorem.
- To prove:
- Perpendicular Bisector Theorem:
- Proof:
- Statement:
- According to perpendicular bisector theorem, in the triangle, any point on perpendicular bisector is at equal distance from both end points of the line segment on which it is drawn.
- Step 1:
Let arbitrary point C on perpendicular bisector.
In which,
CD is perpendicular bisector on AB.
Hence,
AD = DB
CD = CD (common)
So, according to SAS rule
Hence,
CA = CB
So, any point on perpendicular bisector is at equal distance from end points of line segment.
Hence proved.
What is a monomial? Explain with an example.
- We have to define monomial by giving an example.
A monomial is an expression with only one term.
The examples are: 3x, 6y
What is a monomial? Explain with an example.
- We have to define monomial by giving an example.
A monomial is an expression with only one term.
The examples are: 3x, 6y
Find the equation of a line that passes through
and 
We are given two points and we need to find the equation of the line passing through them. The equation of a line passing through two points (a, b) and (c, d) is
Step by step solution:
Let the given points be denoted by
(a, b) = (-3, 1)
(c, d) = (2, -14)
The equation of a line passing through two points (a, b) and (c, d) is
Using the above points, we have
Simplifying the above equation, we have
Cross multiplying, we get
5(y + 14) = -15(x - 2)
Expanding the factors, we have
5y + 70 = -15x + 30
Taking all the terms in the left hand side, we have
15x + 5y + 70 - 30 = 0
Finally, the equation of the line is
15x + 5y + 40 = 0
Dividing the equation throughout by 5, we get
3x + y + 8 = 0
This is the required equation.
Note:
We can simplify the equation in any other way and we would still reach the same equation. The general form of an equation in two variables is given by ax + by + c = 0,, where a, b, c are real numbers. The student is advised to remember all the different forms of a line, like, slope-intercept form, axis-intercept form, etc.
Find the equation of a line that passes through
and 
We are given two points and we need to find the equation of the line passing through them. The equation of a line passing through two points (a, b) and (c, d) is
Step by step solution:
Let the given points be denoted by
(a, b) = (-3, 1)
(c, d) = (2, -14)
The equation of a line passing through two points (a, b) and (c, d) is
Using the above points, we have
Simplifying the above equation, we have
Cross multiplying, we get
5(y + 14) = -15(x - 2)
Expanding the factors, we have
5y + 70 = -15x + 30
Taking all the terms in the left hand side, we have
15x + 5y + 70 - 30 = 0
Finally, the equation of the line is
15x + 5y + 40 = 0
Dividing the equation throughout by 5, we get
3x + y + 8 = 0
This is the required equation.
Note:
We can simplify the equation in any other way and we would still reach the same equation. The general form of an equation in two variables is given by ax + by + c = 0,, where a, b, c are real numbers. The student is advised to remember all the different forms of a line, like, slope-intercept form, axis-intercept form, etc.